This is too big a topic to cover
completely.The first items should be
consulted for older material and the general history.Then I include material of particular interest.See also 6.BL which has some formulae which
are used to computeπ.I have compiled a separate file on the
history ofπ.

Die
Quadratur des Zirkels.Täglicher
Telegraph (Indianapolis) (20 Jan 1897) ??.Surveys attempts since -2000 and notes that Lindemann and Weierstrass
have shown that the problem is impossible, like perpetual motion.

A man of
'genius'.(Indianapolis) Sun (6 Feb
1897) ??.An interview with Goodwin,
who says: "The astronomers have all been wrong.There's about 40,000,000 square miles on the surface of this
earth that isn't here."He says
his results are revelations and gives several rules for the circle and the
sphere.

Squaring the
circle.(Indianapolis) Sunday Journal
(21 Feb 1897) 9.Says Goodwin has
solved all three classical impossible problems.Saysπ = 3.2, using
the fact thatÖ2 = 10/7,giving diagrams and a number of rules.

My thanks to Underwood
Dudley for locating and copying the above newspaper items.

Underwood
Dudley.πt:1832-1879.MM 35 (1962) 153-154.He plots
45 values ofπas a function of time over the period
1832-1879 and finds the least-squares straight line which fits the data,
finding thatπt =
3.14281 + .0000056060 t,fortmeasured in years AD.Deduces
that the Biblical value of 3 was a good approximation for the time and that
Creation must have occurred whenπt
= 0,which was in-560,615.

Underwood
Dudley.πt.JRM 9 (1976-77) 178 & 180.Extends his previous work to 50 values
ofπover 1826-1885, obtainingπt = 4.59183 - .000773 t.The fact thatπt
is decreasing is worrying _ whenπt
= 1,all circles will collapse into
straight lines and this will certainly be the end of the world, which is expected
in 4646 on 9 Aug at 20:55:33 _ though this is only the expected time and there
is considerable variation in this prediction.[Actually, I get that this should be on 11 Aug.However, it seems to me that circles will
collapse onceπt =
2,as then the circumference
corresponds to going back and forth along the diameter.This will occur whent = 3352.949547,i.e. in 3352, on 13 Dec at 14:01:54 _ much earlier than Dudley's
prediction, so start getting ready now!]

6.B. STRAIGHT
LINE LINKAGES

See Yates for a good survey of the
field.

James
Watt.UK Patent 1432 _ Certain New
Improvements upon Fire and Steam Engines, and upon Machines worked or moved by
the same.Granted 28 Apr 1784;complete specification 24 Aug 1784.14pp + 1 plate.Pp. 4-6 & Figures 7‑12 describe Watt's parallel
motion.Yates, below, p. 170 quotes one
of Watt's letters:"... though I
am not over anxious after fame, yet I am more proud of the parallel motion than
of any other invention I have ever made."

A.
Peaucellier.Note sur un balancier
articulé a mouvement rectiligne.Journal de Physique 2 (1873) 388‑390.(Partial English translation in Smith, Source Book, vol. 2,
pp. 324‑325.) Says he communicated it to Soc. Philomath. in 1867 and
that Lipkin has since also found it.There is also an article in Nouv. Annales de Math. (2) 12 (1873) 71‑78
(or 73?), ??NYS.

[J. J.
Sylvester.]Report of the Annual
General Meeting of the London Math. Soc. on 13 Nov 1873.Proc. London Math. Soc. 5 (1873) 4 &
141.On p. 4 is:"Mr. Sylvester then gave a description
of a new instrument for converting circular into general rectilinear motion,
and into motion in conics and higher plane curves, and was warmly applauded at
the close of his address."On p.
141 is an appendix saying that Sylvester spoke "On recent discoveries in
mechanical conversion of motion" to a Friday Evening's Discourse at the
Royal Institution on 23 Jan 1874.It
refers to a paper 20 pages long but is not clear if or where it was published.

R. C.
Archibald.Bibliography of the theory
of linkages.SM 2 (1933‑34) 293‑294.Supplement to Kanayama.

Robert C.
Yates.Geometrical Tools.(As:Tools; Baton Rouge, 1941);revised ed., Educational Publishers, St. Louis, 1949.Pp. 82-101 & 168-191.Gets up to outlining Kempe's proof that any
algebraic curve can be drawn by a linkage.

Item
642: Turning a square by circular motion, p. 247.Plain face, with four pins forming a centred square, is turned by
the lathe.A triangular follower is
against the face, so it is moved in and out as a pin moves against it.This motion is conveyed by levers to the
tool which moves in and out against the work which is driven by the same lathe.

Item
681: Geometrical boring and routing chuck, pp. 257-258.Shows it can make rectangles, triangles,
stars, etc.No explanation of how it
works.

These were discovered by Arthur H.
Stone, an English graduate student at Princeton in 1939.American paper was a bit wider than English
and would not fit into his notebooks, so he trimmed the edge off and had a pile
of long paper strips which he played with and discovered the basic
flexagon.Fellow graduate students
Richard P. Feynman, Bryant Tuckerman and John W. Tukey joined in the
investigation and developed a considerable theory.One of their fathers was a patent attorney and they planned to
patent the idea and began to draw up an application, but the exigencies of the
1940s led to its being put aside, though knowledge of it spread as mathematical
folklore.E.g. Tuckerman's father,
Louis B. Tuckerman, lectured on it at the Westinghouse Science Talent Search in
the mid 1950s.

S&B, pp. 148‑149, show
several versions.Most square versions
(tetraflexagons or magic books) don't fold very far and are really just
extended versions of the Jacob's Ladder _ see 11.L

Martin
Gardner.Cherchez la Femme [magic
trick].Montandon Magic Co., Tulsa,
Okla., 1946.Reproduced in:Martin Gardner Presents; Richard Kaufman and
Alan Greenberg, 1993, pp. 361-363.[In:Martin Gardner Presents, p. 404, this is
attributed to Gardner, but Gardner told me that Roger Montandon had the
copyright _ ??I have learned a little
more about Gardner's early life _ he supported himself by inventing and selling
magic tricks about this time, so it may be that Gardner devised the idea and
sold it to Montandon.].A
hexatetraflexagon.

Sidney
Melmore.A single‑sided doubly
collapsible tessellation.MG 31 (No.
294) (1947) 106.Forms a Möbius strip
of three triangles and three rhombi.He
sees it has two distinct forms, but doesn't see the flexing property!!

William R.
Ransom.A six‑sided hexagon.SSM 52 (1952) 94.Shows how to number the 6 faces.No history.

F. G.
Maunsell.Note 2449:The flexagon and the hexahexaflexagram.MG 38 (No. 325) (Sep 1954) 213‑214.States that Joseph is first article in the
field and that this is first description of the flexagon.Gives inventors' names, but with Tulsey for
Tukey.

Anthony S.
Conrad & Daniel K. Hartline.Flexagons.TR 62-11, RIAS, (7212
Bellona Avenue, Baltimore 12, Maryland,) 1962, 376pp.This began as a Science Fair project in 1956 and was then
expanded into a long report.The
authors were students of Harold V. McIntosh who kindly sent me one of the
remaining copies in 1996.They discover
how to make any chain of polygons into a flexagon, provided certain relations
among angles are satisfied.The
bibliography includes almost all the preceeding items and adds the references
to the Rogers & D'Andrea patent, some other patents (??NYS) and a number of
ephemeral items:Conrad produced an
earlier RIAS report, TR 60-24, in 1960;Allan Phillips wrote a mimeographed paper on hexaflexagons;McIntosh wrote an unpublished paper on
flexagons;Mike Schlesinger wrote an
unpublished paper on Tuckerman tree theory.

Robert E.
Neale (154 Prospect Parkway, Burlington, Vermont, 05401, USA).Self-designing tetraflexagons.12pp document received in 1996 describing
several ways of making tetraflexagons without having to tape or paste.He starts with a creased square sheet, then
makes some internal tears or cuts and then folds things through to miraculously
obtain a flexagon!

6.E.FLEXATUBE

This is the square cylindrical tube
that can be inverted by folding.It was
also invented by Arthur H. Stone, c1939, cf 6.D.

M.
Gardner.Flexa-tube puzzle.Ibidem 7 (Sep 1956) 129.Cites the inventors of the flexagons and the
articles of Maunsell and Leech (but he doesn't have its details).(I have a note that this came with attached
sample, but the copy I have doesn't indicate such.)

John
Fisher.John Fisher's Magic Book.Muller, London, 1968.Homage to Houdini, pp. 152‑155.Detailed diagrams of the solution, but no
history.

Highland
Games (2 Harpers Court, Dingwall, Ross-Shire, IV15 9HT) makes a version called
Table Teaser, made in a strip with end pieces magnetic.Pieces are coloured so to produce several
folding and inverting problems other than the usual one.Bought in 1995.

NOTATION.Each of the types of puzzle considered has a basic unit and
pieces are formed from a number of these units joined edge to edge.The notationN: n1, n2, ....denotes a puzzle withNpieces, of whichnipieces consist ofibasic units.Ifni are
single digit numbers, the intervening commas and spaces will be omitted, but
the digits will be grouped by fives, e.g.15: 00382 11.

G. P.
Jellis.Special Issue on Chessboard
Dissections.Chessics 28 (Winter 1986)
137‑152.Discusses many problems
and early work in Fairy Chess Review.

Branko
Grünbaum & Geoffrey C. Shephard.Tilings and Patterns.Freeman,
1987.Section 9.4: Polyiamonds,
polyominoes and polyhexes, pp. 497-511.Good outline of the field with a number of references otherwise unknown.

Michael
Keller.A polyform timeline.World Game Review 9 (Dec 1989)
4-5.This outlines the history of
polyominoes and other polyshapes.Keller and others refer to polyoboloes as polytans.

Rodolfo
Marcelo Kurchan (Parana 960 5 "A", 1017 Buenos Aires,
Argentina).Puzzle Fun, starting with
No. 1 (Oct 1994).This is a magazine
entirely devoted to polyomino and other polyform puzzles.Many of the classic problems are extended in
many ways here.In No. 6 (Aug 1995) he
presents a labelling of the 12 hexiamonds by the lettersA, C, H, I, J, M, O, P, S, V, X, Y,which he obtained from Anton Hanegraaf.I have never seen this before.

Hooper.Rational Recreations.Op. cit. in 4.A.1.1774.Vol. 1, recreation
23, pp. 64-66.Considers figures formed
of isosceles right triangles.He has
eight of these, coloured with eight colours, and uses some of them to form
"chequers or regular four-sided figures, different either in form or
colour".

Book of 500
Puzzles.1859.Triangular problem, pp. 74-75.Identical to Hooper, dropping the last
sentence.

John Milner
Lester.US Patent 1,290,761 _ Game
Apparatus.Filed 6 Feb 1918;patented 7 Jan 1919.2pp + 3pp diagrams.Fairly general assembly puzzle claims.He specifically illustrates a polyomino
puzzle and a polyabolo puzzle.The
first has a Greek cross of edge3(hence containing45unit cells) to be
filled with polyominoes _11: 01154.The second has
an8-pointed star formed by
superimposing two4 x 4squares.This has area20and hence contains40isosceles right
triangles of edge1, which is the basic
unit of this type of puzzle.There
are11: 0128pieces.

Blyth.Match-Stick Magic.1921.Spots and squares,
pp. 68-73.He uses matchsticks broken
in thirds, so it is easier to describe with units of one-third.6 units,4 doubles and2 triples.Some of the pieces have black bands or
spots.Object is to form polyomino
shapes without pieces crossing, but every intersection must have a black
spot.19polyomino shapes are given to construct, including7of
the pentominoes, though some of the shapes are only connected at corners.

"John
Bull" Star of Fortune Prize Puzzle.1922.This is a puzzle with20pieces, coloured red on one side, containing6through13triangles to be assembled into a star of David with4triangles along each edge (hence12 x 16 = 192triangles).Made by Chad Valley.Prize of £250 for a red star matching the
key solution deposited at a bank;£150
for solution closest to the key;£100
for a solution with10red and10grey pieces, or as nearly as
possible.Closing date of competition
is 27 Dec 1922.Puzzle made by Chad
Valley Co. as a promotional item for John Bull magazine, published by
Odhams Press.A copy is in the toy shop
of the Buckleys Shop Museum, Battle, East Sussex, to whom I am indebted for the
chance to examine the puzzle and a xerox of the puzzle, box and solution.

Daily Sketch
Jig-Saw Puzzle.By Chad Valley.Card polyiamonds.39: 0,0,1,5,6, 12,9,6,with a path printed on one side, to assemble into a shape of16rows of15with four corners removed and so the printed
sides form a continuous circuit.In box
with shaped bottom.Instructions on
inside cover and loose sheet to submit solution.No dates given, but appears to be 1920s, though it is somewhat
similar to the Daily Mail Crown Puzzle of 1953 _ cf below _ so it might be much
later.

B. T.s
Zig-Zag.B.T. is a Copenhagen
newspaper.Polyiamond puzzle.33: 0,0,1,2,5 6,7,2,2,4 1,1,1,Some repetitions, so I only see 20 different
shapes.To be fit into an irregular
frame.Solution given on 23 Nov 1931,
pp. 1-2.(I have a photocopy of the
form to fill in; an undated set of rules, apparently from the paper, saying the
solutions must be received by 21 Nov; and the pages giving the solution;
provided by Jan de Geus.)

F.
Kadner.Solution 1597.Problemist Fairy Chess Supplement (later
called Fairy Chess Review) 2:10 (Feb 1935) 104-105.Shows the35hexominoes cannot tile a rectangle by two
arguments, both essentially based on two colouring.Gives some other results and some problems are given as 1679-1681
_ ??NYS.

William E.
Lester.Correction to 1597.Problemist Fairy Chess Supplement (later
called Fairy Chess Review) 2:11 (Apr 1935) 121.Corrects an error in Kadner.Finds a number of near-solutions.Editor says Kadner insists the editor should take credit for the
two-colouring form of the previous proof.

H. D.
Benjamin, proposer.Problem 3228.Fairy Chess Review 3:12 (Jun 1938) 129.Dissect a5 x 5into the five
tetrominoes and a pentomino so that the pentomino touches all the tetrominoes
along an edge.Asserts the solution is
unique.Refers to problems 3026‑3030
_ ??NYS.

H. D.
Benjamin, proposer.Problem 3229.Fairy Chess Review 3:12 (Jun 1938) 129.Dissect an8 x 8into the12pentominoes and a tetromino so that all pieces touch the edge of the
board.Asserts only one tetromino
works.

T. R.
D[awson], proposer.Problems
3230-1.Fairy Chess Review 3:12 (Jun
1938) 129.Extends prob. 3229 to ask
for solutions with12pieces on the edge, using two other
tetrominoes.Thinks it cannot be done
with the remaining two tetrominoes.

J.
Niemann.Item 4154: "The colossal
count".Fairy Chess Review 4:3
(Nov 1939) 44-45.Announces that there
are3698-ominoes,12859-ominoes and465410-ominoes, but
Keller and Jelliss note that he missed a10‑omino which was not corrected until 1966.

H. D.
Benjamin.Unpublished notes.??NYS _ cited and briefly described in G. P.
Jelliss; Prob. 48 _ Aztec tetrasticks; G&PJ 2 (No. 17) (Oct 1999) 320.Jelliss says Benjamin studied polysticks,
which he called 'lattice dissections' around 1946-1948 and that some results by
him and T. R. Dawson were entered in W. Stead's notebooks but nothing is known
to have been published.For orders1, 2, 3, 4,there are1, 2, 5, 16polysticks.Benjamin formed these into a6 x
6lattice square.Jelliss then mentions Barwell's rediscovery
of them and goes on to a new problem _ see Knuth, 1999.

R.
J. French.Space dissections.Fairy Chess Review 7:2 (Oct 1948) 16
(erroneously printed as 108).French
writes that he and A. W. Baillie have corrected the number of6-cubes to35 + 77 + 54 = 166.Baillie notes that every6-cube lies in two layers _ i.e. has some
width£ 2_ and asks for the result forn‑cubes as prob. 7879.[I
suspect the answer is thatn £
3kimplies that ann-cube has some width£
k.]Editor adds some corrections to the
discussion in 6:18.

Daily Mail
Crown Puzzle.Made by Chad Valley
Co.1953.26 pieces, coloured on one side, to be fit into a crown shape.11 are border pieces and easily placed.The other 15 are polyiamonds:15: 00112 24012 11.Prize of £100 for solution plus best slogan,
entries due on 8 Jun 1953.

S. W.
Golomb.Checkerboards and
polyominoes.AMM 61 (1954) 675‑682.Mostly concerned with covering the8 x 8board with copies of polyominoes.Shows one covering with the12pentominoes and the square
tetromino.Mentions that the idea can
be extended to hexagons.S&B, p.
18, and Gardner (Dec 1964) say he mentions triangles, but he doesn't.

Jules
Pestieau.US Patent 2,900,190 _
Scientific puzzle.Filed 2 Jul
1956;patented 18 Aug 1959.2pp + 1p diagrams.For the12pentominoes!Diagram shows the6 x
10solution with two5 x 6rectangles and shows the two-piece non-symmetric equivalence of theNandFpieces.Pieces have
markings on one side which may be used _ i.e. pieces may not be turned
over.Mentions possibility of
usingn-ominoes.

Gardner.SA (Dec 1957) = 1st Book, chap. 13.Exposits Golomb and Stead.Gives number ofn-ominoes forn = 1, ...,
7.1st Book describes Scott's
work.Says a pentomino set called
'Hexed' was marketed in 1957.(John
Brillhart gave me and my housemates an example in 1960 _ it took us two weeks
to find our first solution.)

Dana
Scott.Programming a Combinatorial
Puzzle.Technical Report No. 1, Dept.
of Elec. Eng., Princeton Univ., 1958, 20pp.Uses MANIAC to find65solutions for pentominoes on an8 x 8board with square2 x 2in the centre.Notes that the3 x 20pentomino rectangle has just two
solutions.In 1999, Knuth notes that
the total number of solutions with the2 x 2being anywhere does not
seem to have ever been published and he finds16146.

M.
Gardner.SA (Sep 1958) c= 2nd Book,
chap. 6.First general mention of solid
pentominoes, pentacubes, tetracubes.In
the Addendum in 2nd Book, he says Theodore Katsanis of Seattle suggested the
eight tetracubes and the29pentacubes in a letter to Gardner on
23 Sep 1957.He also says that
Julia Robinson and Charles W. Stephenson both suggested the solid pentominoes.

C. Dudley
Langford.Note 2793:A conundrum for form VI.MG 42 (No. 342) (Dec 1958) 287.4each of theL,N,andT (= Y)tetrominoes make a7 x 7square with the
centre missing.Also nine pieces make
a6 x 6square but this requires an even number ofTs.

C. B. &
Jenifer Haselgrove.A computer program
for pentominoes.Ibid., 16‑18.Outlines program which found the2339solutions for the6 x 10.It is usually said that they also found all
solutions of the3 x 20,4 x 15and5 x 12,but I don't see it mentioned here and in JRM
7:3 (1974) 257, it is reported that Jenifer (Haselgrove) Leech stated that only
the6 x 10and3 x 20were donein 1960, but that she did the5
x 12and4 x 15with a new program
in c1966.See Fairbairn, c1962, and
Meeus, 1973.

T. H.
O'Beirne.Pentominoes and
hexiamonds.New Scientist 12 (No. 259)
(2 Nov 1961) 316‑317.This is the
first use of the word 'polyiamond'.He
considers the19one‑sided pieces.He says he devised the pieces and R. K. Guy
has already published many solutions in Nabla.He asks for the number of ways the 18 one-sided pentominoes can fill
a9 x 10.In 1999, Knuth found this would take several months.

M.
Gardner.Polyiamonds.SA (Dec 1964) = 6th Book, chap. 18.Exposits basic ideas and results for the 12
double sided hexiamonds.Poses several
problems which are answered by readers.The six-pointed star using 8 pieces has a unique solution.John G. Fletcher and Jenifer (Haselgrove)
Leech both showed the3 x 12rhombus is impossible.Fletcher found the3 x 11rhombus has24solutions, all omitting the 'bat'.Leech found155solutions for the6 x 6rhombus and74solutions for the 4 x 9.Mentions there are1609-iamonds, one with a
hole.

John G.
Fletcher.A program to solve the
pentomino problem by the recursive use of macros.Comm. ACM 8 (1965) 621-623.??NYS _ described by Knuth in 1999 who says that Fletcher found the 2339
solutions for the6 x 10in 10 minutes on an IBM 7094 and that the
program remains the fastest known method for problems of placing the 12
pentominoes.

Joseph S.
Madachy.Pentominoes _ Some solved and
unsolved problems.JRM 2:3 (Jul 1969)
181-188.Gives the numbers of Parkin,
Lander & Parkin.Shows various
examples where a rectangle splits into two congruent halves.Discusses various other problems, including
Bouwkamp's3 x 4 x 5solid pentomino problem.Bouwkamp reports that the final total of
3940 was completed on 16 Mar 1967 after about three years work using three
different computers, but that a colleague's program would now do the whole
search in about three hours.

P. J.
Torbijn.Polyiamonds.JRM 2:4 (Oct 1969) 216-227.Uses the double sided hexiamonds and
heptiamonds.A few years before, he
found, by hand, that there are156ways to cover the6 x 6rhombus with the 12
hexiamonds and74ways for the4 x 9,but could find no
way to cover the3 x 12.The previous year, John G. Fletcher
confirmed these results with a computer and he displays all of these _ but this
contradicts Gardner (Dec 64) _ ??He
gives several other problems and results, including using the 24 heptiamonds to
form7 x 12,6 x 14,4 x 21and3 x 28rhombuses.

Solomon W.
Golomb.Tlling with sets of
polyominoes.J. Combinatorial Theory 9
(1970) 60‑71.??NYS.Extends his 1966 paper.Asks which heptominoes tile rectangles and
says there are two undecided cases _ cf. Marlow, 1985.Gardner (Aug 75) says Golomb shows that the
problem of determining whether a given finite set of polyominoes will tile the
plane is undecidable.

M.
Gardner.SA (Sep 1972).c= Knotted, chap. 3.Says the8tetracubes were made by
E. S. Lowe Co. in Hong Kong and marketed as "Wit's
End".Says an MIT group found1390solutions for the2 x 4 x 4box packed with tetracubes.He reports that several people found that
there are1023heptacubes _ but see Niemann, 1948,
above.Klarner reports that the
heptacubes fill a2 x 6 x 83.

Says he drew out all the solutions for
the area 60 rectangles in 1972 (cf Fairbairn, c1962).Finds that520of the6 x 10rectangles can be divided
into two congruent halves, sometimes in two different ways.For5 x 12,there are380;for4 x 15,there are 94.Gives some hexomino rectangles by either deleting a piece or
duplicating one, and an 'almost11 x
19'.Says there are46solutions to the3 x 30with the 18 one-sided pentominoes and
attributes this to Mrs (Haselgrove) Leech, but the correction indicates this
was found by A. Mank.

Jenifer
Haselgrove.Packing a square with
Y-pentominoes.JRM 7:3 (1974) 229.She finds and shows a way to pack 45
Y-pentominoes into a15 x 15,but is unsure if there are more solutions.In 1999, Knuth found212solutions.She also reports the
impossibility of using the Y-pentominoes to fill various other rectangles.

S. W.
Golomb.Trademark for
'PENTOMINOES'.US trademark 1,008,964
issued 15 Apr 1975;published
21 Jan 1975 as SN 435,448.(First
use:November 1953.)[These appear in the Official Gazette of the
United States Patent Office (later Patent and Trademark Office) in the
Trademarks section.]

Anon.31: Polyominoes.QARCH 1:8 (June 1984) 11‑13. [This is an occasional publication of The Archimedeans, the
student maths society at Cambridge.]Good survey of counting and asymptotics for the numbers of polyominoes,
up ton = 24,polycubes, etc.10 references.

Bernard
Wiezorke & Jacques Haubrich.Dr.
Dragon's polycons.CFF 33 (Feb 1994)
6-7.Polycons (for connections) are the
same as the polysticks described by Barwell in 1990, above.Authors describe a Taiwanese version on sale
in late 1993, using10of the4‑sticks suitably shortened so they fit into the grooves of a4 x 4board _ so crossings are not permitted.(Ann x nboard hasn+1lines ofnedges in each direction.)They
fit15of the4-sticks onto a5 x 5board and determine all solutions.

CFF 35 (Dec 1994) 4 gives a
number of responses to the article.Brain Barwell wrote that he devised them as a student at Oxford, c1970,
but did not publish until 1990.He expected
someone to say it had been done before, but no one has done so.He also considered using the triangular and
hexagonal lattice.He had just
completed a program to consider fitting15of the 4-sticks onto a5 x 5board and found over180,000solutions, with slightly
under half having no crossings, confirming the results of Wiezorke &
Haubrich.

Dario Uri also wrote that he
had invented the idea in 1984 and called them polilati (polyedges).Giovanni Ravesi wrote about them in Contromossa
(Nov 1984) 23 _ a defunct magazine.

Chris
Roothart.Polylambdas.CFF 34 (Oct 1994) 26-28.A lambda is a30o-60o-90otriangle.These may be joined along corresponding legs, but not along
hypotenuses.Forn = 1, 2, 3, 4, 5,there are1, 4, 4, 11,
12n-lambdas.He gives some problems using various sets of these pieces.

Richard
Guy.Letters of 29 May and 13 Jun
1996.He is interested in using
the19one-sided hexiamonds.Hexagonal
rings of hexagons contain1, 6, 12hexagons, so the hexagon with three hexagons
on a side has 19 hexagons.If these
hexagons are considered to comprise six equilateral triangles, we have a board
with19 x 6triangles.O'Beirne asked
for the number of ways to fill this board with the one-sided hexiamonds.Guy has collected over4200solutions.A program by Marc Paulhus
found907solutions in eight hours, from which it initially estimated that
there are about30,000solutions.The second letter gives the final results _ there are124,518solutions.This is modulo
the12symmetries of the hexagon.In
1999, Knuth found124,519 and Paulhus has rerun his program and found
this number.

Hilarie
Korman.Pentominoes: A first player
win.IN: Games of No Chance; ed. by
Richard Nowakowski; CUP, 1997??, ??NYS - described inWilliam Hartston; What mathematicians get up to; The Independent
Long Weekend (29 Mar 1997) 2.This
studies the game proposed by Golomb _ players alternately place one of the
pentominoes on the chess board, aligned with the squares and not overlapping
the previous pieces, with the last one able to play being the winner.She used a Sun IPC Sparcstation for five
days, examining about22 x 109positions to show the game is a first player
win.

Nob
Yoshigahara found in 1994 that the smallest box which can be packed with
W-pentacubes is5 x 6 x 6.In 1997, Yoshya (Wolf) Shindo found that one
can pack the6 x 10 x 10with Z-pentacubes, but it is not known if
this is the smallest such box.These
were the last unsolved problems as to whether a box could be packed with a
planar pentacube (= solid pentomino).

Marcel Gillen&Georges Philippe.Twinform462 Puzzles in one.Solutions for Gillen's puzzle exchange at
17IPP, 1997, 32pp + covers.Take 6 of
the pentominos and place them in a7 x
5rectangle, then place the other six
to make the same shape on top of the first shape.There are462(= BC(12,6)/2)possible puzzles and all of them have solutions.TakingF, T, U, W, X, Zfor the first
layer, there is just one solution; all other cases have multiple solutions,
totalling22,873solutions, but only one solution for each
case is given here.)

Richard K.
Guy.O'Beirne's hexiamond.In:The Mathemagican and Pied Puzzler; ed. by Elwyn Berlekamp & Tom
Rodgers, A. K. Peters, Natick, Massachusetts, 1999, pp. 85‑96.He relates that O'Beirne discovered the 19
one-sided hexiamonds in c1959 and found they would fill a hexagonal shape in
Nov 1959 and in Jan 1960 he found a solution with the hexagonal piece in the
centre.He gives Paulhus's results (see
Guy's letters of 1996), broken down in various ways.He gives the number of double-sided (i.e. one can turn them over)
and single-sidedn-iamonds forn = 1, ..., 7.Cf Ellard, 1982, for many more values for the double-sided case.

n1234567

double111341224

single111461944

In 1963, Conway and Mike Guy
considered looking for 'symmetric' solutions for filling the hexagonal shape
with the 19 one-sided hexiamonds.A
number of these are described.

Donald E.
Knuth.Dancing links.25pp preprint of a talk given at Oxford in
Sep 1999, sent by the author.Available
as:http://www-cs-faculty.stanford.edu/~knuth/preprints.html .In this he introduces a new technique for
backtrack programming which runs faster (although it takes more storage) and is
fairly easy to adapt to different problems.In this approach, there is a symmetry between pieces and cells.He applies it to several polyshape problems,
obtaining new, or at least unknown, results.He extends Scott's 1958 results to get16146ways to pack the8 x 8with the 12 pentominoes and the2 x 2.He describes
Fletcher's 1965 work.He extends
Haselgrove's 1973 work and finds 212 ways to fit 15 Y-pentominoes in a15 x 15.Describes Torbijn's 1969 work and Paulhus' 1996 work on hexiamonds,
correcting the latter's number to124,519.He then looks for the
most symmetric solutions for filling the hexagonal shape with the 19 one-sided
hexiamonds, in the sense discussed by Guy (1999).He then considers the 18 one-sided pentominoes (cf Meeus (1973))
and tries the9 x 10,but finds it would take a few months on his
computer (a 500 MHz Pentium III), so he's abandoned it for now.He then considers polysticks, citing an
actual puzzle version that I've not seen.He adapts his program to them.He considers the 'welded tetrasticks' which have internal junction
points.There are six of these and ten
if they are taken as one-sided.The ten
can be placed in a4 x 4 grid.There are15unwelded, one-sided,
tetrasticks, but they do not form a square, nor indeed any nice shape.He considers all25one-sided tetrasticks
and asks if they can be fit into what he calls an Aztec Diamond, which is the
shape looking like a square tilted 45o on the square lattice.The rows contain1, 3, 5, 7, 9, 7, 5, 3, 1cells.He thinks an exhaustive
search is beyond present computing power.

G. P.
Jelliss.Prob. 48 _ Aztec
tetrasticks.G&PJ 2 (No. 17) (Oct
1999) 320.Jelliss first discusses
Benjamin's work on polysticks (see at 1946-1948 above) and Barwell's
rediscovery of them (see above).He
then describes Knuth's Dancing Links and gives the Aztec Diamond problem.Jelliss has managed to get all but one of
the polysticks into the shape, but feels it is impossible to get them all in.

6.F.1. OTHER CHESSBOARD DISSECTIONS

See S&B, pp. 12‑14.See also 6.F.5 for dissections of uncoloured
boards.

Jerry
Slocum.Compendium of Checkerboard
Puzzles.Published by the author,
1983.Outlines the history and shows
all manufactured versions known then to him:33 types in 61 versions.The
first number in Slocum's numbers is the number of pieces.

Jerry
Slocum&Jacques Haubrich.Compendium of Checkerboard Puzzles.2nd ed., published by Slocum, 1993.90 types in 161 versions, with a table of which pieces are in which
puzzles, making it much easier to see if a given puzzle is in the list or
not.This gives many more pictures of
the puzzle boxes and also gives the number of solutions for each puzzle and
sometimes prints all of them.The
Slocum numbers are revised in the 2nd ed. and I use the 2nd ed. numbers below.(There was a 3rd ed. in 1997, NYR. Les Barton is working on an extended version)

??UK patent application 16,810.1892.Not granted, so never published.??It may be possible to see the application??(Hordern has an example with this number on it, by Feltham &
Co.However, in the 2nd ed., the cover
is reproduced and it looks like the number may be 16,310.)14: 00149.Slocum 14.20.1.Manufactured as: The Chequers Puzzle, by Feltham & Co.

Hoffmann.1893.Chap. III, no. 16: The chequers puzzle, pp. 97‑98 & 129‑130.14: 00149.Slocum 14.20.1.Says it is made
by Messrs Feltham, who state it has over 50 solutions.He gives two solutions.(Photo in Hordern, p. 61.)

Western
Puzzle Works, 1926 Catalogue.No. 79:
"Checker Board Puzzle, in 16 pieces", but the picture only shows 14
pieces.14: 00149.Picture doesn't show any colours, but
assuming the standard colouring of a chess board, this is the same as Slocum
14.15.

Emil
Huber-Stockar.L'echiquier du
diable.Comptes-Rendus du Deuxième
Congrès International de Récréation Mathématique, Paris, 1937.Librairie du "Sphinx", Bruxelles,
1937, pp. 64-68.Discusses how one
solution can lead to many others by partial symmetries.Shows several solutions containing about 40
altogether.Note at end says he has now
got5275solutions.This article
is reproduced in Sphinx 8 (1938) 36-41, but without the extra pages of diagrams.At the end, a note says he has5330solutions.Ibid, pp. 75-76 says
he has got 5362 solutions and ibid. 91-92 says he has5365.By use of Bayes'
theorem on the frequency of new solutions, he estimatesc5500solutions.Haubrich has found 6013.He intended to produce a book of solutions,
but he died in May 1939 [Sphinx 9 (1939) 97].

F.
Hansson.Sam Loyd's 18-piece dissection
_ Art. 48 & probs. 4152‑4153.Fairy Chess Review 4:3 (Nov 1939) 44.Cites Loyd's Puzzles Magazine.Asserts there are many millions of solutions!He determines the number of chequered handedn-ominoes forn = 1, 2, ..., 8is2, 1, 4, 10, 36, 110, 392,
1371.The first 17 pieces total 56
squares.Considers 8 ways to dissect
the board into 18 different pieces.Problems
ask for the number of ways to choose the pieces in each of these ways and for
symmetrical solutions.Solution in 4:6
(Jun 1940) 93-94 (??NX of p. 94) says there are a total of3,309,579ways to make the choices.

C. Dudley
Langford.Note 2864:A chess‑board puzzle.MG 43 (No. 345) (Oct 1959) 200.15: 01248.Example with the underside coloured with reversed colours.Gives two solutions and says there is at
least one more.Not in Slocum.

Leonard J.
Gordon.Broken chessboards with unique
solutions.G&PJ 10 (1989) 152‑153.Shows Dudeney's problem has four solutions.Finds other colourings which give only one
solution.Notes some equivalences in
Slocum.

6.F.2.COVERING
DELETED CHESSBOARD WITH DOMINOES

See also 6.U.2.

There is nothing on this in
Murray.

Pál Révész.Op. cit. in 5.I.1.1969.On p. 22, he says
this problem comes from John [von] Neumann, but gives no details.

R. E.
Gomory.(Solution for deletion of any
two squares of opposite colour.)In:M. Gardner, SA (Nov 1962) =
Unexpected, pp. 186‑187.Solution
based on a rook's tour.(I don't know
if this was ever published elsewhere.)

Michael
Holt.What is the New Maths?Anthony Blond, London, 1967.Pp. 68 & 97.Gives the4 x 4case as a
problem, but doesn't mention that it works on other boards.(I include this as I haven't seen earlier examples
in the educational literature.)

David
Singmaster.Covering deleted
chessboards with dominoes.MM 48 (1975)
59‑66.Optimum extension ton‑dimensions.For ann-dimensional board, each dimension must be³
2.If the board has an even number of
cells, then one can delete anyn-1white cells and anyn-1black cells and still cover the board with dominoes (i.e.2 x 1 x 1 x ... x 1blocks).If the board has an odd number of cells, then let the corner cells be
coloured black.One can then delete
anynblack cells and anyn-1white cells and still cover the board with
dominoes.

I-Ping Chu
& Richard Johnsonbaugh.Tiling
deficient boards with trominoes.MM
59:1 (1986) 34-40.(3,n) = 1andn ¹
5imply that ann x nboard with one cell deleted can be covered withLtrominoes.Some5 x 5boards with one cell deleted can be tiled, but not all can.

6.F.3.DISSECTING
A CROSS INTOZsANDLs

Minguét.Engaños.1733.Pp. 119-121 (1755: 85-86;
1822: 138-139).Cross into 5 pieces,
similar to Les Amusemens, but oneZis longer and oneLis
shorter.Diagram shows8LandZshaped pieces, but it
is not clear what the next problem wants _ either a piece or a label is
missing.Says one can make different
figures with the pieces.

Bestelmeier.1801.Item 274 _ Das mathematische Kreuz.Cross into 6 pieces, but the picture has an erroneous extra line.It should be the reversal of the picture in
Catel, the same as in Les Amusemens.

Charles
Babbage.The Philosophy of Analysis _
unpublished collection of MSS in the BM as Add. MS 37202, c1820.??NX.See 4.B.1 for more details.F. 4
is "Analysis of the Essay of Games".F. 4.v has the dissection of the cross into 3Zpentominos and twoLpieces.

Vyse.Tutor's Guide.1771?Prob. 9, p. 317
& Key p. 358.Refers to the land as
a parallelogram though it is drawn rectangular.

Charles
Babbage.The Philosophy of Analysis _
unpublished collection of MSS in the BM as Add. MS 37202, c1820.??NX.See 4.B.1 for more details. F. 4
is "Analysis of the Essay of Games".F. 4.v has an entry"8½
aProb of figure"followed by theL‑tromino.8½ b is
the same with a mitre and there are other dissection problems adjacent _ see
6.F.3, 6.AQ, 6.AW.1, 6.AY, so it seems clear that he knew this problem.

Prob.
21: Puzzle of the four tenants, pp. 235 & 260.Identical to Magician's Own Book.

Prob.
27: Puzzle of the two fathers, pp. 237‑238 & 262.Identical to Magician's Own Book.

Illustrated
Boy's Own Treasury.1860.Prob. 21, pp. 399 & 439.15/16of a square to be divided into five (congruent) parts, each with two
trees.c= Magician's Own Book,
prob. 3.

Leske.Illustriertes Spielbuch für Mädchen.1864?Prob. 175, p. 88.L-tromino into
four congruent pieces, each with two trees.The problem is given in terms of the original square to be divided into
five parts, where the father gets a quarter of the whole in the form of a
square and the four sons get congruent pieces.

Loyd.Origin of a famous puzzle _ No. 18: An
ancient puzzle.Tit‑Bits 31 (13
Feb&6 Mar 1897) 363&419.Nearly 50 years ago he was told of the
quadrisection of3/4of a square, but drew the mitre shape
instead of theL‑tromino.See 6.AW.1.

Benson.1904.The farmer's puzzle, p. 196.Quadrisect anL‑tromino.

Wehman.New Book of 200 Puzzles.1908.

The
divided garden, p. 17.= Magician's Own
Book, prob. 3

Puzzle
of the two fathers, p. 43.= Magician's
Own Book, prob. 28.

Puzzle
of the four tenants, p. 46.=
Magician's Own Book, prob. 22.

Dudeney.Some much‑discussed puzzles.Op. cit. in 2.1908.Land in shape of
anL‑tromino to be
quadrisected.He says this is supposed
to have been invented by Lord Chelmsford (Sir F. Thesiger), who died in 1878 _
see Charades, Enigmas, and Riddles (1859?).But cf Les Amusemens.

Rowan
Barnes-Murphy.Monstrous
Mysteries.Piccolo, 1982.Apple-eating monsters, pp. 40 & 63.Trisect into equal parts, the shape
consisting of a2 x 4rectangle with a1 x 1square attached to
one of the central squares of the long side.[Actually, this can be done with the square attached to any of the
squares, though if it as attached to the end of the long side, the resulting
pieces are straight trominoes.]

Family
Friend 3 (1850) 90 & 121.Practical
puzzle _ No. XIII.4 x 4square, with 12 trees in the corners,
centres of sides and four at the centre of the square, to be divided into 4
congruent parts each with 3 trees.Solution uses 4L-trominoes.The same problem is
repeated as Puzzle 17 _ Twelve-hole puzzle in (1855) 339 with solution in
(1856) 28.

P. Hein, et
al.Soma booklet.Parker Bros., 1969, 56pp.Asserts there are240simple solutions
and1,105,920total solutions, found by J. H. Conway & M. J. T. Guy with a a
computer (but cf Gardner, below) and by several others.[There seem to be several versions of this
booklet, of various sizes.]

Thomas V.
Atwater, ed.Soma Addict.4 issues, 1970‑1971, produced by
Parker Brothers.(Gardner, below, says
only three issues appeared.)??NYS _
can anyone provide a set or xeroxes??

M.
Gardner.SA (Sep 1972) c= Knotted,
chap. 3.States there are240solutions for the cube, obtained by many programs, but first found by J.
H. Conway & M. J. T. Guy in 1962, who did not use a computer, but did it by
hand "one wet afternoon".

See also 6.N, 6.U.2, 6.AY.1 and
6.BJ.The predecessors of these puzzles
seem to be the binomial and trinomial cubes showing(a+b)3and(a+b+c)3.I have an example of the latter from the
late 19C.Here I will consider only
cuts parallel to the cube faces _ cubes with cuts at angles to the faces are in
6.BJ.Most of the problems here involve
several types of piece _ see 6.U.2 for packing with one kind of piece.

M.
Gardner.SA (Sep 1972) c= Knotted,
chap. 3.Discusses Hoffmann's
Diabolical Cube and Mikusi_ski's cube.Says he has 8 solutions for the first and that there are just2for the second.The Addendum
reports that Wade E. Philpott showed there are just13solutions of the
Diabolical Cube.Conway has confirmed
this.Gardner briefly describes the
solutions.Gardner also shows the Lesk
Cube, designed by Lesk Kokay (Mathematical Digest [New Zealand] 58 (1978)
??NYS), which has at least3solutions.

Leisure
Dynamics, the US distributor of Impuzzables, a series of63
x 3 x 3cube dissections identified by
colours, writes that they were invented by Robert Beck, Custom Concepts Inc.,
Minneapolis.However, the Addendum to
Gardner, above, says they were designed by Gerard D'Arcey.

Anon. [= John
Leech, according to Gardner, below].Two dissection problems:2.Eureka 13 (1950) 6&14 (1951) 23.Asks for such a dissection using at most 10
pieces.Gives an 8 piece solution due
to R. F. Wheeler.[Cundy & Rollett;
Mathematical Models; 2nd ed., pp. 203‑205, say Eureka is the first
appearance they know of this problem.See Gardner, below, for the identity of Leech.]

M.
Gardner.SA (Oct 1973) c= Knotted,
chap. 16.He says that the problem was
posed by John Leech.He gives Wheeler's
initials as E. H. ??He says that
J. H. Thewlis found a simpler 8‑piece solution, further simplified by T.
H. O'Beirne, which keeps the4 x 4 x 4cube intact.This is shown in
Gardner.Gardner also shows an 8‑piece
solution which keeps the5 x 5 x 5intact, due to E. J. Duffy, 1970.O'Beirne showed that an 8‑piece
dissection into blocks is impossible and found a 9‑block solution in
1971, also shown in Gardner.

Richard K.
Guy, proposer;editors & Charles H.
Jepson [should be Jepsen], partial solvers.Problem 1122.CM 12 (1987)
50&13 (1987) 197‑198.Asks for such dissections under various conditions, of which (b) is the
form given in Eureka.Eight pieces is
minimal in one case and seems minimal in two other cases.Eleven pieces is best known for the first
case, where the pieces must be blocks, but this appears to be the problem
solved by O'Beirne in 1971, reported in Gardner, above.

David
Singmaster.Comment on the
"Spots" puzzle.29 Sep 1994,
2pp.Letter in response to the above.I note that there is no standard pattern for
a die other than the opposite sides adding to seven.There are23 =
8ways to orient the spots forming2, 3, and 6.There are two handednesses, so there are 16 dice altogether.(This was pointed out to me perhaps 10 years
before by Richard Guy and Ray Bathke.I
have since collected examples of all 16 dice.)However, Ray Bathke showed me Oriental dice with the two spots of the 2
placed horizontal or vertically rather than diagonally, giving another 16 dice
(I have 5 types), making 32 dice in all.A die can be dissected into91 x 1 x 3pieces in 6 ways if the layers have to alternate in direction, or in 21
ways in general.I then pose a number
of questions about such dissections.

6.G.4.USE
OF OTHER POLYHEDRAL PIECES

S&B.1986.P. 42 shows Stewart Coffin's 'Pyramid Puzzle' using pieces made from
truncated octahedra and his 'Setting Hen' using pieces made from rhombic
dodecahedra.Coffin probably devised
these in the 1960s _ perhaps his book has some details of the origins of these
ideas.??check.

Mark Owen
& Matthew Richards.A song of six
splats.Eureka 47 (1987) 53‑58.There are six ways to join three truncated
octahedra.For reasons unknown, these
are called 3‑splats.They give
various shapes which can and which cannot be constructed from the six 3‑splats.

6.H.PICK'S
THEOREM

Georg
Pick.Geometrisches zur
Zahlenlehre.Sitzungsberichte des
deutschen naturwissenschaftlich‑medicinischen Vereines für Böhmen
"Lotos" in Prag (NS) 19 (1899) 311‑319.Pp. 311‑314 gives the proof, for an
oblique lattice.Pp. 318‑319
gives the extension to multiply connected and separated regions.Rest relates to number theory.[I have made a translation of the material
on Pick's Theorem.]

Charles
Howard Hinton.The Fourth
Dimension.Swan Sonnenschein & Co.,
London, 1906.Metageometry, pp. 46-60.[This material is in Speculations on the
Fourth Dimension, ed. by R. v. B. Rucker; Dover, 1980, pp. 130-141.Rucker says the book was published in 1904,
so my copy may be a reprint??]In the
beginning of this section, he draws quadrilateral shapes on the square lattice
and determines the area by counting points, but he countsI + E/2 + C/4,which works for quadrilaterals but is not valid in general.

H.
Steinhaus.O mierzeniu pól
p_askich.Przegl_d Matematyczno‑Fizyczny
2 (1924) 24‑29.Gives a version
of Pick's theorem, but doesn't cite Pick.(My thanks to A. M_kowski for an English summary of this.)

Richard A.
Gibbs.Pick iff Euler.MM 49 (1976) 158.Cites DeTemple & Robertson and observes that both Pick and
Euler can be proven from a result on triangulations.

John
Reay.Areas of hex-lattice polygons,
with short sides.Abstracts Amer. Math.
Soc. 8:2 (1987) 174, #832-51-55.Gives
a formula for the area in terms of the boundary and interior points and the
characteristic of the boundary, but it is an open question to determine when
this formula gives the actual area.

6.I.SYLVESTER'S
PROBLEM OF COLLINEAR POINTS

If a set of non‑collinear points
in the plane is such that the line through any two points of the set contains a
third point of the set, then the set is infinite.

G. A.
Dirac.Note 2271:On a property of circles.MG 36 (No. 315) (Feb 1952) 53‑54.Replace 'line' by 'circle' in the problem.He shows this is true by inversion.He asks for an independent proof of the
result, even for the case whentwo,
threeare replaced bythree, four.

Carlile.Collection.1793.Prob. CV, p. 62.A dog and a duck are in a circular pond of
radius 40 and the swim at the same speed.The duck is at the edge and swims around the circumference.The dog starts at the centre and always
toward the duck, so the dog and the duck are always on a radius.How far does the dog swim in catching the
duck.He simply gives the result
as20π.LettingRbe the radius of the pond andVbe
the common speed, I find the radius of the dog,r,is given byr = R sin Vt/R.Since the angle,θ,of both the duck and the
dog is given byθ = Vt/R,the polar equation of the dog's path isr = R sin θand the path is a semicircle whose diameter is the appropriate
radius perpendicular to the radius to the duck's initial position.

Cambridge
Math. Tripos examination, 5 Jan 1871, 9 to 12.Problem 16, set by R. K. Miller.Three bugs in general position, but with velocities adjusted to make
paths similar and keep the triangle similar to the original.

Pearson.1907.Part II, no. 66: A duck hunt, pp. 66 & 172.Duck swims around edge of pond;spaniel starts for it from the centre at the
same speed.

A. S.
Hathaway, proposer and solver.Problem
2801.AMM 27 (1920) 31&28 (1921) 93‑97.Pursuit
of a prey moving on a circle.Morley's
and other solutions fail to deal with the case when the velocities are
equal.Hathaway resolves this and shows
the prey is then not caught.

Nelson F.
Beeler & Franklyn M. Branley.Experiments in Optical Illusion.Ill. by Fred H. Lyon.Crowell,
1951, An illusion doodle, pp. 68-71, describes the pattern formed by four bugs
starting at the corners of a square, drawing the lines of sight at
(approximately) regular intervals.Putting several of the squares together, usually with alternating
directions of motion, gives a pleasant pattern which is now fairly common.They call this 'Huddy's Doodle', but give no
source.

J. E.
Littlewood.A Mathematician's
Miscellany.Op. cit. in 5.C.1953.'Lion and man', pp. 135‑136 (114‑117).The 1986 ed. adds three diagrams and revises
the text somewhat.I quote from
it."A lion and a man in a closed
circular arena have equal maximum speeds.What tactics should the lion employ to be sure of his meal?"This was "invented by R. Rado in the
late thirties" and "swept the country 25 years later".[The 1953 ed., says Rado didn't publish
it.]The correct solution "was
discovered by Professor A. S. Besicovitch in 1952".[The 1953 ed. says "This has just been
discovered ...; here is the first (and only) version in print."]

C. C.
Puckette.The curve of pursuit.MG 37 (No. 322) (Dec 1953) 256‑260.Gives the history from Bouguer in 1732.Solves a variant of the problem.

I. J.
Good.Pursuit curves and mathematical
art.MG 43 (No. 343) (Feb 1959) 34‑35.Draws tangent to the pursuit curves in an
equilateral triangle and constructs various patterns with them.Says a similar but much simpler pattern was
given by G. B. Robison; Doodles; AMM 61 (1954) 381-386, but Robison's doodles
are not related to pursuit curves, though they may have inspired Good to use
the pursuit curves.

J. Charles
Clapham.Playful mice.RMM 10 (Aug 1962) 6‑7.Easy derivation of the distance traveled
fornbugs at corners of a regularn‑gon.[I don't see this result in Bernhart.]

C. G.
Paradine.Note 3108:Pursuit curves.MG 48 (No. 366) (Dec 1964) 437‑439.Says Good makes an error in Note 3079.He shows the length of the pursuit curve in
the equilateral triangle is_of the side and describes the curve as an
equiangular spiral.Gives a simple
proof that the length of the pursuit curve in the regularn‑gon is the side divided by(1 ‑ cos 2π/n).

M. S.
Klamkin & D. J. Newman.Cyclic
pursuit or "The three bugs problem".AMM 78 (1971) 631‑639.General treatment.Cites
Bernhart's four SM papers and some of the history therein.

Western
Puzzle Works, 1926 Catalogue.No. 1712
_ unnamed, but shows both the square and the triangle.Apparently a four piece puzzle.

Adams.Puzzle Book.1939.Prob. C.153:
Squaring a triangle, pp. 162 & 189.Asserts that Dudeney's method works for any triangle, but his example is
close to equilateral and I recall that this has been studied and only certain
shapes will work??

Robert C.
Yates.Geometrical Tools.(As:Tools; Baton Rouge, 1941);revised ed., Educational Publishers, St. Louis, 1949.Pp. 40-41.Extends to dissecting a quadrilateral to a specified triangle and gives
a number of related problems.

6.L.CROSSED
LADDERS

Two ladders are placed across a
street, each reaching from the base of the house on one side to the house on
the other side.

The simple problem gives the
heightsa,bthat the ladders reach
on the walls.If the height of the crossing
isc,we easily get1/c = 1/a +
1/b.NOTATION _ this problem will be
denoted by(a, b).

The more common and more complex
problem is where the ladders have lengthsaandb,the height of their
crossing iscand one wants the widthdof the street.If the heights of the ladder ends arex,y,this situation givesx2 ‑ y2 = a2
‑ b2and1/x + 1/y = 1/cwhich leads to a quartic and there seems to be no simple
solution.NOTATION _ this will be
denoted(a, b, c).

Mahavira.850.Chap. VII, v. 180-183, pp. 243-244.Gives the simple version with a modification _ each ladder reaches from
the top of a pillar beyond the foot of the other pillar.The ladder from the top of pillarY(of heighty)extends bymbeyond the foot of pillarXand the ladder from the top of pillarX(of heightx)reachesnbeyond the foot of pillarY.The pillars aredapart.Similar triangles then yield:(d+m+n)/c=(d+n)/x + (d+m)/yand one can compute the various distances along the ground.He first does problems withm = n = 0,which are the simple version of the problem, but sincedis
given, he also asks for the distances on the ground.

Fibonacci.1202.Pp. 397‑398 looks like a crossed ladders problem but is a simple
right triangle problem.

Pacioli.Summa.1494.Part 2, f. 56r, prob.
48.(4, 6).

Hutton.A Course of Mathematics.1798?Prob. VIII,1833: 430;1857: 508.A ladder40long in a roadway can reach33up one side and, from the same point, can reach21up the other side.This is
actually a simple right triangle problem.

Loyd.Problem 48: A jubilee problem.Tit‑Bits 32 (21 Aug,11&25 Sep 1897) 385,439&475.Given heights of the ladder ends above
ground and the width of the street, find the height of the intersection.However one wall is tilted and the drawing
has it covered in decoration so one may interpret the tilt in the wrong way.

J. S.
Cromelin, proposer;Murray Barbour,
solver.Problem E616 _ The three
ladders.AMM 51 (1944) 231&592.Ladders of length60&77from one side.A ladder from the other side crosses them at heights17&19.How long is the third ladder and how wide is
the street?

Anon.A window cleaner's problem.Mathematical Pie 51 (May 1967) 399.From a point in the road, a ladder can reach30ft up on one side and40ft up on the other side.If the two ladder positions are at right
angles, how wide is the road?

A ladder of lengthLis
placed to just clear a box of widthwand heighthat
the base of a wall.How high does the
ladder reach?Denote this by(w, h, L).Lettingxbe the horizontal distance of the foot andybe
the vertical distance of the top of the ladder, measured from the foot of the
wall, we getx2 + y2
= L2and(x‑w)(y‑h) = wh,which gives a quartic in general.But ifw = h,then use ofx + yas a variable reduces the quartic to a quadratic.In this case, the idea is old _ see e.g.
Simpson.

The question of determining shortest
ladder which can fit over a box of widthwand heighthis
the same as determining the longest ladder which will pass from a corridor of
widthwinto another corridor of widthh.See Huntington below and
section 6.AG.

Simpson.Algebra.1745.Section XVIII, prob. XV,
p. 250 (1790: prob. XIX, pp. 272-273)."The Side of the inscribed SquareBEDF,and the HypotenuseACof a right-angled TriangleABCbeing given; to determine the
other two Sides of the TriangleABandBC."Solves "by
consideringx + yas one Quantity".

D. John
Baylis.The box and ladder
problem.MTg 54 (1971) 24.(2, 2, 10).Finds the quartic which he solves by symmetry.Editorial note in MTg 57 (1971) 13 says several people wrote to
say that use of similar triangles avoids the quartic.

A. A.
Huntington.More on ladders.M500 145 (Jul 1995) 2-5.Does usual problem, getting a quartic.Then finds the shortest ladder.[This turns out to be the same as the
longest ladder one can get around a corner from corridors of widthswandh,so this problem is connected to 6.AG.]

These involve finding the shortest
distance over the surface of a cube or cylinder.I've just added the cylindrical form _ see Dudeney (1926),
Perelman and Singmaster.I don't know
if other shapes have been done _ the regular (and other) polyhedra and the cone
could be considered.

Round
the cone, pp. 144 & 195.What is
the shortest distance from a pointParound a cone and back toP?Answer is "An ellipse", which doesn't seem to answer the
question.If the cone has heightH,radiusRandPislfrom the apex,then the slant heightLisÖ(R2 + H2),the angle of the opened out cone isθ = 2πR/Land the required distance is2l sin θ/2.

Spider
circuit, pp. 144 & 198.Spider is
at the midpoint of an edge of a cube.He wants to walk on each of the faces and return.What is his shortest route?Answer is "A regular hexagon.(This may be demonstrated by putting a
rubber band around a cube.)"

David
Singmaster.The spider spied her.Problem used as:More than one way to catch a fly, The Weekend Telegraph (2 Apr
1984).Spider inside a glass tube, open
at both ends, goes directly toward a fly on the outside.When are there two equally short paths?Can there be more than two shortest routes?

Michael
Goldberg.A duplication of the cube by
dissection and a hinged linkage.MG 50
(No. 373) (Oct 1966) 304‑305.Shows that a hinged version exists with 10 pieces.Hanegraaf, below, notes that there are
actually 12 pieces here.

The projection of a unit cube along a
space diagonal is a regular hexagon of sideÖ2/Ö3.The largest square inscribable in this
hexgon has edgeÖ6 - Ö2 =
1.03527618.By passing the larger cube
on a slant to the space diagonal, one can get the larger cube having edge3Ö2/4 = 1.06066172.

Cundy and
Rollett, p. 158, give references to Zacharias (see below) and to Cantor, but
Cantor only cites Hennessy.

H.
Hennessy.Ronayne's cubes.Phil. Mag. (5) 39 (Jan‑Jun 1895) 183‑187.Quotes, from Gibson's 'History of Cork', a
passage taken from Smith's 'History of Cork', 1st ed., 1750, vol. 1, p. 172,
saying that Philip Ronayne had invented this and that a Daniel Voster had made
an example, which may be the example owned by Hennessy.He finds the dimensions.

W. A.
Bagley.Puzzle Pie.Op. cit. in 5.D.5.1944.No. 12: Curios
[sic] cubes, p. 14.First says it can
be done with equal cubes and then a larger can pass through a smaller.Claims that the larger cube can be about1.1,but this is due to an error _ he thinks the hexagon has the same
diameter as the cube itself.

(W.
Leybourn.Pleasure with Profit.London, 1694.??I cannot recall the
source of this reference and think it may be an error.I have examined the book and find nothing
relevant in it.)

Loyd.Cyclopedia, 1914, pp. 288 & 378.8 x 8to5 x 13and to an area of63.Asserts Loyd
presented the first of these in 1858.Cf. Loyd Jr, below.

O.
Schlömilch.Ein geometrisches
Paradoxon.Z. Math. Phys. 13 (1868)
162.8 x 8to5 x 13.(This article is only signed Schl.Weaver, below, says this is Schlömilch, and
this seems right as he was a co‑editor at the time.Coxeter (SM 19 (1953) 135‑143) says it
is V. Schlegel, apparently confusing it with the article below.)Doesn't give any explanation, leaving it as
a student exercise.

Richard A.
Proctor.Some puzzles.Knowledge 9 (Aug 1886) 305-306."We suppose all the readers ... know
this old puzzle."Describes and
explains8 x 8to5
x 13.Gives a different method of
cutting so that each rectangle has half the error _ several typographical
errors.

W.
Ahrens.Mathematische Spiele.Teubner, Leipzig.3rd ed., 1916, pp. 94‑95 & 111‑112.The 4th ed., 1919, and 5th ed., 1927, are
identical with the 3rd ed., but on different pages:pp. 101‑102&pp. 118‑119.Art. X.65 = 64 = 63gives8 x 8to5 x 13and to area63.The area63case does not appear in the 2nd ed., 1911, which has Art. V.64 = 65,pp. 107 & 118‑119 and this material is not in the 1st
ed. of 1907.

Tom
Tit??In Knott, 1918, but I can't find
it in Tom Tit.No. 3: The square and
the rectangle:64 = 65!,pp. 15-16.Clearly explained.

Collins.Book of Puzzles.1927.A paradoxical
puzzle, pp. 4-5.8 x 8to5
x 13.Shades the unit cells that the lines
pass through and sees that one way has 16 cells, the other way has 17 cells,
but gives only a vague explanation.

Schwenter.1636.Discusses Serlio's dissection and observes the area change.[??Schwenter has not yet been entered.]

I have a
vague reference to the 1723 ed. of Ozanam, but I have not seen it in the 1725
ed. _ this may be an error for the 1778 ed. below.

Vyse.Tutor's Guide.1771?Prob. 8, p. 317
& Key p. 358.Lady has a table27square and a board12 x 48.She cuts the board into two12 x 24rectangles and cuts each rectangle along a diagonal.By placing the diagonals of these pieces on
the sides of her table, she makes a table36square.Note that362 = 1296and272 + 12 x 24=1305.Vyse is clearly unaware
that area has been created.By dividing
all lengths by3,one gets a version where one unit of area is
lost.Note that4, 8, 9is almost a Pythagorean triple.

William
Hooper.Rational Recreations.1774.Op. cit. in 4.A.1.Vol. 4, pp.
286‑287: Recreation CVI _ The geometric money.3 x 10cut into four
pieces which make a2 x 6and a4 x 5.(The diagram is shown in Gardner, MM&M, pp. 131‑132.)(I recently saw that an edition erroneously
has a3 x 6instead of a2 x 6rectangle.This must be the 1st ed. of 1774, as it is correct in my 2nd ed. of
1782.)

Loyd.Get Off the Earth.Puzzle notices in the Brooklyn Daily Eagle (26 Apr ‑ 3 May
1896), printing individual Chinamen.Presenting all of these at an office of the newspaper gets you an
example of the puzzle.Loyd ran
discussions on it in his Sunday columns until 3 Jan 1897 and he also sold
many versions as advertising promotions.S&B, p. 144, shows several versions.

Dudeney.The world's best puzzles.Op. cit. in 2.1908.Gives "Get Off
the Earth" on p. 785.

Loyd.Teddy and the Lions.Gardner, MM&M, p. 123, says he has seen
only one example, made as a promotional item for the Eden Musee in
Manhattan.This has a round disc, but
two sets of figures _ 7 natives and 7 lions which become 6 natives and 8 lions.

W. A.
Bagley.Puzzle Pie.Op. cit. in 5.D.5.1944.No. 24: A
chessboard fallacy, pp. 28-29.8 x
8cut with a diagonal of a8 x 7region, then pieces slid and a triangle cut off and moved to the other
end to make a9 x 7.Clear illustration.

Mel
Stover.From 1951, he devised a number
of variations of both Get off the Earth (perhaps the best is his Vanishing
Leprechaun) and of Teddy and the Lions (6 men and 4 glasses of beer become 5
men and 5 glasses).I have examples of
some of these from Stover and I have looked at his notebooks.SeeGardner, MM&M, pp. 125-128.

John
Fisher.John Fisher's Magic Book.Muller, London, 1968.

Financial
Wizardry, pp. 18-19.7 x 8region with£signs marking the area.A line cuts off a triangle of width 7 and
height 2 at the top.The rest of the
area is divided by a vertical into strips of widths 4 and 3, with a small
rectangle 3 by 1 cut from the bottom of the width 3 strip.When the strips are exchanged, one unit of area
is lost and one£sign has vanished.

Dean
Clark.A centennial tribute to Sam
Loyd.CMJ 23:5 (Nov 1992) 402‑404.Gives an easy circular version with11 & 12astronauts around the earth and a15 & 16face version with
three pieces, a bit like the Vanishing Leprechaun.

Vol.
2, 1892.L'Étoile à cinq branches, pp.
153-154.= K, no. 5: The pentagon and
the five pointed star, pp. 20‑21.He adds that folding over the free end and holding the knot up to the
light shows the pentagram.

E.
Fourrey.Procédés Originaux de
Constructions Géométriques.Vuibert,
Paris, 1924.Pp. 113 & 135‑139.Cites Lucas and cites d'Aviso as Traitè de
la Sphère and says he gives the pentagonal and hexagonal knots.Fourrey shows and describes both, also
giving the pictures on his title page.

(An earlier version of the
book, by Lietzmann & Trier, appeared in 1913, with 2nd ed. in 1917.The 3rd ed. of 1923 was divided into two
books:Wo Steckt der Fehler?andTrugschlüsse.There was a 4th
ed. in 1937.The relevant material
would be in Trugschlüsse, but I have not seen any of the relevant books, though
Northrop cites Lietzmann, 1923, three times _ ??NYS.)

S. L.
Tabachnikov.Errors in geometrical
proofs.Quantum 9:2 (Nov/Dec 1998)
37-39 & 49.Shows:every triangle is isosceles (6.R.1);the sum of the angles of a triangle is 180o
without use of the parallel postulate;a rectangle inscribed in a square is a square;certain approaching lines never meet (6.R.3);all circles have the same circumference (cf
Aristotle's Wheel Paradox in 10.A.1);the circumference of a wheel is twice its radius;the area of a sphere of radiusRisπ2R2.

6.R.1.EVERY
TRIANGLE IS ISOSCELES

This is
sometimes claimed to have been in Euclid's lost Pseudaria.

Ball.MRE, 1st ed., 1892, pp. 33‑34.On p. 32, Ball refers to Euclid's lost
Fallacies and presents this fallacy and the one in 6.R.2:"I do not know whether either of them
has been published previously."In
the 3rd ed., 1896, pp. 42-43, he adds the heading:To prove that every triangle is isosceles.In the 5th ed., 1911, p. 45, he adds a note
that he believes these two were first published in his 1st ed. and notes that
Carroll was fascinated by them and they appear in The Lewis Carroll Picture
Book _ see below.

W. A.
Bagley.Puzzle Pie.Op. cit. in 5.D.5.1944.Call Mr. Euclid _
No. 16: To prove one right angle greater than another right angle, pp.
18-19."Here again, if you take
the trouble to draw an accurate diagram, you will find that the
"construction" used for the alleged proof is impossible."

E. A.
Maxwell.Note 2121:That every angle is a right angle.MG 34 (No. 307) (Feb 1950) 56‑57.Detailed demonstration of the error.

6.R.3.LINES
APPROACHING BUT NOT MEETING

Proclus.5C.A Commentary on the First Book of Euclid's Elements.Translated by Glenn R. Morrow.Princeton Univ. Press, 1970.Pp. 289-291.Gives the argument and tries to refute it.

Jeremy
Gray.Ideas of Space.OUP, 1979.Pp. 37-39 discusses Proclus' arguments in the context of attempts to
prove the parallel postulate.

6.R.4.OTHERS

Ball.MRE, 3rd ed, 1896, pp. 44-45.To prove that, if two opposite sides of a
quadrilateral are equal, the other two sides must be parallel.Cites Mathesis (2) 3 (Oct 1893) 224 _ ??NYS

Cecil B.
Read.Mathematical fallacies&More mathematical fallacies.SSM
33 (1933) 575‑589&977-983.There are two perpendiculars from a point to a line.Part of a line is equal to the whole
line.Every triangle is isosceles (uses
trigonometry).Angle trisection (uses a
marked straightedge).

Ronald C.
Read.Tangrams _ 330 Puzzles.Dover, 1965.The Introduction, pp. 1-6, is a sketch of the history.Will Shortz says this is the first serious attempt
to counteract the mythology created by Loyd and passed on by Dudeney.Read cannot get back before the early 1800s
and notes that most of the Loyd myth is historically unreasonable.However, Read does not pursue the early
1800s history in depth and I consider van der Waals to be the first really
serious attempt at a history of the subject.

Jan van der
Waals.History&Bibliography.In:Joost Elffers; Tangram; (1973), Penguin,
1976.Pp. 9‑27&29‑31.Says the Chinese
term "ch'i ch'ae" dates from the Chu era (‑740/‑330), but
the earliest known Chinese book is 1813.The History reproduces many pages from early works.The Bibliography cites 8 versions of 4
Chinese books (with locations!) from 1813 to 1826 and 18 Western books from
1805 to c1850.

Hoffmann.1893.Chap III, pp. 74‑144.Describes Tangrams and Richter puzzles at some length on pp. 74‑90.Lots of photos in Hordern.

Recent
research by Jerry Slocum, backed up by The Admired Chinese Puzzle, indicates
that the introduction of tangrams into Europe was done by a person or persons
in Lord Amherst's 1815-1817 embassy to China, which visited Napoleon on St. Helena
on its return voyage.If so, then the
conjectural dating of several items below needs to be amended.I have amended my discussion accordingly and
marked such dates with ??.Although
watermarking of paper with the correct date was a legal requirement at the
time, paper might have been stored for some time before it was printed on, so
watermark dates only give a lower bound for the date of printing.I have seen several further items dated
1817, but it is conceivable that some material may have been sent back to
Europe a few years earlier.

SPECIFIC ITEMS

Kanchusen.Wakoku Chiekurabe.1727.Pp. 9 & 28-29:
a simple dissection puzzle with 8 pieces.The problem appears to consist of a mitre comprising¾of
a unit square;4 isoceles right
triangles of hypotenuse 1and3 isoceles right triangles of side½,but the solution shows that all the triangles are the same size, say
having hypotenuse1,and the mitre shape is actually formed from
a rectangle of size1 x Ö2.

"Ganriken"
[pseud., possibly of Fan Chu Sen].Sei
Sh_nagon Chie-no-Ita (The Ingenious Pieces by Sei Sh_nagon.) (In
Japanese).Kyoto Shobo, Aug 1742, 18pp,
42 problems and solutions.Reproduced
in a booklet, ed. by Kazuo Hanasaki, Tokyo, 1984, as pp. 19‑36. Also reproduced in a booklet, transcribed
into modern Japanese, with English pattern names and an English abstract, by
Shigeo Takagi, 1989.This uses a set of
seven pieces different than the Tangram.S&B, p. 22, shows these pieces.Sei Sh_nagon (c965-c1010) was a famous courtier, author of The Pillow
Book and renowned for her intelligence.The Introduction is signed Ganriken.S&B say this is probably Fan Chu Sen, but Takagi says the author's
real name is unknown.

Utamaro.Interior of an Edo house, from the picture‑book:The Edo Sparrows (or Chattering Guide),
1786.Reproduced in B&W in:J. Hillier; Utamaro _ Colour Prints and
Paintings; Phaidon Press, Oxford, (1961), 2nd ed., 1979, p. 27, fig. 15.I found this while hunting for the next item.This shows two women contemplating some
pieces but it is hard to tell if it is a tangram‑type puzzle, or if
perhaps they are cakes.Hiroko and Mike
Dean tell me that they are indeed cooking cakes.

Utamaro.Woodcut.1792.Shows two courtesans working
on a tangram puzzle.van der Waals
dated this as 1780, but Slocum has finally located it, though he has only been
able to find two copies of it!The
courtesans are clearly doing a tangram-like puzzle with 12(?) pieces _ the
pieces are a bit piled up and one must note that one of the courtesans is
holding a piece.They are looking at a
sheet with 10 problem figures on it.

A New
Invented Chinese Puzzle, Consisting of Seven Pieces of Ivory or Wood, Viz.
5 Triangles, 1 Rhomboid, & 1 Square, which may be so placed as to form
the Figures represented in the plate.Paine & Simpson, Boro'.Undated, but the paper is watermarked 1806.This consists of two 'volumes' of 8 pages each, comprising 159
problems with no solutions.At the end
are bound in a few more pages with additional problems drawn in _ these are direct
copies of plates 21, 26, 22, 24, and 28 (with two repeats from plate 22) of The
New and Fashionable Chinese Puzzle, 1817.Bound in plain covers.This is
in Edward Hordern's collection and he has provided a photocopy.

Ch'i Ch'iao
t'u ho‑pi (Harmoniously combined book of tangram problems).1813.(Bibliothek Leiden 6891, with an 1815 edition at British Library 15257 d
13.)??NYS (van der Waals).323 examples.The 1813 seems to be the earliest Chinese tangram book of
problems, with the 1815 being the solutions.Slocum says there was a solution book in 1815 and that the problem book
had a preface bySang‑hsia
K'o,which was repeated in the solution
book with the same date.A version of
this appears to have been the book given to Napoleon and to have started the
tangram craze in Europe.I have a
version from c1820s which has 334 problems.

Shichi‑kou‑zu
Gappeki [The Collection of Seven‑Piece Clever Figures].Hobunkoku Publishing, Tokyo, 1881.This is a Japanese translation of an 1813
Chinese book "recognized as the earliest of existing Tangram book",
apparently the previous item.[The book
says 1803, but Jerry Slocum reports this is an error for 1813!]Reprinted, with English annotations by Y.
Katagiri, from N. Takashima's copy, 1989.129 problems (but he counts 128 because he omits one after no. 124), all
included in my version of the previous item, no solutions.

A Key to the
New and Fashionable Chinese Puzzle, Published by J. and E. Wallis, 42, Skinner
Street, London, Wherein is explained the method of forming every Figure
contained in That Pleasing Amusement.Nd [Mar 1817].PHOTOCOPY from
the Bodleian Library, Oxford, catalogue number Jessel e.1176.TP seems to made by pasting in the cover
slip and has been bound in as a left hand page.ALSO a PHOTOCOPY from Jerry Slocum.In the latter copy, the apparent TP appears to be a paste down on
the cover.The latter copy does not
have the Stanzas mentioned below.Slocum's copy has paper watermarked 1815; I didn't check this at the
Bodleian.

NOTE.This is quite a different book than The New
and Fashionable Chinese Puzzle published by Goodrich in New York, 1817.

Bound in at the beginning of
the Fashionable Chinese Puzzle and the Bodleian copy of the Key is:Stanzas, Addressed to Messrs. Wallis, on the
Ingenious Chinese Puzzle, Sold by them at the Juvenile Repository, 42, Skinner
Street.In the Key, this is on
different paper than the rest of the booklet.The Stanzas has a footnote referring to the ex-Emperor Napoleon as being
in a debilitated state.(Napoleon died
in 1821, which probably led to the Bodleian catalogue's date of c1820 for the
entire booklet - but see below.Then
follow 28 plates with 323 numbered figures (but number 205 is skipped), solved
in the Key.In the Bodleian copy of the
Key, these are printed on stiff paper, on one side of each sheet, but arranged
as facing pairs, like Chinese booklets.

[Philip A. H. Brown; London
Publishers and Printersc. 1800-1870;
British Library, 1982, p. 212] says the Wallis firm is only known to have
published under the imprint J. & E. Wallis during 1813 and Ruth Wallis
showed me another source giving 1813?-1814.This led me to believe that the booklets originally appeared in 1813 or
1814, but that later issues or some owner inserted the c1820 sheet of Stanzas,
which was later bound in and led the Bodleian to date the whole booklet as
c1820.Ruth Wallis showed me a source
that states that John Wallis (Jun.) set up independently of his father at 186
Strand in 1806 and later moved to Sidmouth.Finding when he moved to Sidmouth might help date this publication more
precisely, but it may be a later reissue.However, Slocum has now found the book advertised in the London Times in
Mar 1817 and says this is the earliest Western publication of tangrams, based
on the 1813/1815 Chinese work.Wallis
also produced a second book of problems of his own invention and some copies
seem to be coloured.

In AM, p. 43, Dudeney says
he acquired the copy ofThe Fashionable
Chinese Puzzlewhich had belonged to
Lewis Carroll.He says it was "Published
by J. and E. Wallis, 42 Skinner Street, and J. Wallis, Jun., Marine Library,
Sidmouth" and quotes the Napoleon footnote, so this copy had the Stanzas
included.This copy is not in the
Strens Collection at Calgary which has some of Dudeney's papers.

Van der Waals cites two
other works titledThe Fashionable
Chinese Puzzle.An 1818 edition from A.
T. Goodridge [sic], NY, is in the American Antiquarian Society Library (see
below) and another, with no details given, is in the New York Public Library.Could the latter be the Carroll/Dudeney
copy?

Toole Stott 823 is a copy
with the same title and imprint as the Carroll/Dudeney copy, but he dates it
c1840.This version is in two
parts.Part I has 1 leaf text + 26 col.
plates _ it seems clear that col. means coloured, a feature that is not
mentioned in any other description of this book _ perhaps these were
hand-coloured by an owner.Unfortunately, he doesn't give the number of puzzles.I wonder if the last two plates are missing
from this??Part II has 1 leaf text +
32 col. plates, giving 252 additional figures.The only copy cited was in the library of J. B. Findlay _ I have
recently bought a copy of the Findlay sale catalogue, ??NYR.

Toole Stott 1309 is listed
with the title: Stanzas, ....J. &
F. [sic] Wallis ... and Marine Library, Sidmouth, nd [c1815].This has 1 leaf text and 28 plates of
puzzles, so it appears that the Stanzas have been bound in and the original
cover title slip is lost or was not recognised by Toole Stott.The date of c1815 is clearly derived from
the Napoleon footnote but 1817 would have been more reasonable, though this may
be a later reissue.Again only one copy
is cited, in the library of Leslie Robert Cole.

Plates 1-28 are identical to
plates 1-28 of The Admired Chinese Puzzle, but in different order.The presence of the Chinese text in The
Admired Chinese Puzzle made me think the Wallis version was later than it.

Comparison of the Bodleian
booklet with the first 27 plates of Giuoco Cinese, 1818?, reveals strong
similarities.5 plates are essentially
identical, 17 plates are identical except for one, two or three changes and 3
plates are about 50% identical.I find
that 264 of the 322 figures in the Wallis booklet occur inGiuoco Cinese,which is about 82%.However, even when the plates are essentially identical, there are often
small changes in the drawings or the layout.

Some of the plates were
copied by hand into Hordern's copy ofA
New Invented Chinese Puzzle, c1806??.

The Admired
Chinese PuzzleA New & Correct
Edition From the Genuine Chinese Copy.C. Taylor, Chester, nd [1817].Paper is clearly watermarked 1812, but the Prologue refers to the book
being brought from China by someone in Lord Amherst's embassy to China, which
took place in 1815-1817 and which visited Napoleon on St. Helena on its
return.Slocum dates this to after 17
Aug 1817, when Amherst's mission returned to England and this seems to be the
second western book on tangrams.Not in
Christopher, Hall, Heyl or Toole Stott _ Slocum says there is only one copy
known in England!It originally had a cover
with an illustration of two Chinese, titledThe Chinese Puzzle,and one of
the men holds a scroll sayingTo amuse
and instruct.The Chinese text gives
the titleCh'i ch'iao t'u ho pi
(Harmoniously combined book of tangram problems).I have a photocopy of the cover from Slocum.Prologue facing TP; TP; two pp in Chinese,
printed upside down, showing the pieces;32pp of plates numbered at the upper left (sometimes with reversed
numbers), with problems labelled in Chinese, but most of the characters are
upside down!The plates are printed
with two facing plates alternating with two facing blank pages.Plate 1 has 12 problems, with solution lines
lightly indicated.Plates 2 - 28
contain 310 problems.Plates 29-32
contain 18 additional "caricature Designs" probably intended to be
artistic versions of some of the abstract tangram figures.The Prologue shows faint guide lines for the
lettering, but these appear to be printed, so perhaps it was a quickly done
copperplate.The text of the Prologue
is as follows.

This
ingenious geometrical Puzzle was introduced into this Kingdom from China.

The following sheets are a
correct Copy from the Chinese Publication, brought to England by a Gentleman of
high Rank in the suit [sic] of Lord Amherst's late Embassy.To which are added caricature Designs as an
illustration, every figure being emblematical of some Being or Article known to
the Chinese.

The plates are identical to the plates
in The Fashionable Chinese Puzzle above, but in different order and plate 4 is
inverted and this version is clearly upside down.

Sy
Hall.A New Chinese Puzzle,The Above Consists of Seven Pieces of Ivory
or Wood, viz. 5 Triangles, 1 Rhomboid, and 1 Square, which will form the 292
Characters, contained in this Book; Observing the Seven pieces must be used to
form each Character.NB.This Edition has been corrected in all its
angles, with great care and attention.Engraved by Sy Hall, 14 Bury Street, Bloomsbury.Watermarked 1815.31 plates with 292 problems.Slocum and Hordern have copies.Sy probably is an abbreviation of Sydney (or possibly
Stanley?).(I have seen a copy in the
BL, bound with a large folding Plate 2 by Hall, which has 83 examples with
solution lines drawn in (by hand??), possibly one of four sheets giving all the
problems in the book.However there is
no relationship between the Plate and the book _ problems are randomly placed
and often drawn in different orientation.I have a photocopy of the plate on two A3 sheets.)

A New
Chinese Puzzle.Third Edition:
Universally allowed to be the most correct that has been published.1817.Dalgety has a copy.

Miss
Lowry.A Key to the Only Correct
Chinese Puzzle Which has Yet Been Published, with above a Hundred New
Figures.No. 1.Drawn and engraved by Miss Lowry.Printed by J. Barfield, London, 1817.Hordern has a copy.

W.
Williams.New Mathematical
Demonstrations of Euclid rendered clear and familiar to the minds of youth,
with no other mathematical instruments than the triangular pieces, commonly
called the Chinese Puzzle.London,
1817.??NYS (van der Waals).

The
New and Fashionable Chinese Puzzle.A.
T. Goodrich & Co., New York, 1817.TP, 1p of Stanzas (seems like there should be a second page??), 32pp
with 346 problems.Slocum has a copy.

[Key]
to the Chinese Philosophical Amusements.A. T. Goodrich & Co., New York, 1817.TP, 2pp of stanzas (the second page has the Napoleon footnote and
a comment which indicates it is identical to the material in the problem book),
Index to the Key to the Chinese Puzzle, 80pp of solutions as black shapes with
white spacing.Slocum has a copy.

NOTE.This is quite a different book than The
Fashionable Chinese Puzzle published in London by Wallis in 1817.

Slocum writes:
"Although the Goodrich problem book has the same title as the British book
by Wallis and Goodrich has the "Stanzas" poem (except for the first 2
paragraphs which he deleted) the problem books have completely different layouts
and Goodrich's solution book largely copies Chinese books."

Anon.Buonapartes Geliefkoosste Vermaack op St.
Helena, op Chineesch Raadsel.1er
Rotterdam by J. Harcke.Prijs 1 - 4 ??.2e Druck te(?) Rotterdam.Ter Steendrukkery van F. Scheffers &
Co.Nanco Bordewijk has recently
acquired this and Slocum has said it is a translation of one of the English
items in c1818.I have just a copy of
the cover, and it uses many fancy letters which I don't guarantee to have read
correctly.

Giuoco
CineseOssiaRaccolta di 364. Figure Geometrica [last letter is blurred]
formate con un Quadrato diviso in 7. pezzi, colli quali si ponno formare
infinite Figure diversi, come Vuomini[sic], Bestie, Ucelli[sic], Case, Cocchi,
Barche, Urne, Vasi, ed altre suppelletili domestiche: Aggiuntovi l'Alfabeto, e
li Numeri Arabi, ed altre nuove Figure.Agapito Franzetti alle Convertite, Rome, nd [but 1818 is written in by
hand].Copy at the Warburg Institute,
shelf mark FMH 4050.TP & 30
plates.It has alternate openings
blank, apparently to allow you to draw in your solutions, as an owner has done
in a few cases.The first plate shows
the solutions with dotted lines, otherwise there are no solutions.There is no other text than on the TP,
except for a florid headingAlfabetoon plate XXVIII.The diagrams have no numbers or names.The upper part of the TP is a plate of three
men, intended to be Orientals, in a tent?The one on the left is standing and cutting a card marked with the
pieces.The man on the right is sitting
at a low table and playing with the pieces.He is seated on a box labelledROMPI CAPO.A third man is
seated behind the table and watching the other seated man.On the ground are a ruler, dividers and
right angle.The Warburg does not know
who put the date 1818 in the book, but the book has a purchase note showing it
was bought in 1913.James Dalgety has
the only other copy known.Sotheby's
told him that Franzetti was most active about 1790, but Slocum finds Sotheby's
is no longer very definite about this.I thought it possible that a page was missing at the beginning which
gave a different form of the title, but Dalgety's copy is identical to this
one.The letters and numbers are quite
different to those shown in Elffers and the other early works that I have seen,
but there are great similarities to The New and Fashionable Chinese Puzzle,
c1813, and some similarities to Al Gioco Cinese above.I haven't counted the figures to verify the
364.

Bestelmeier,
1823.Item 1278: Chinese Squares.It is not in the 1812 catalogue.

Slocum.Compendium.Shows the above Bestelmeier entry.

Anonymous.Ch'i ch'iao t'u ho pi (Harmoniously combined
book of Tangram problems)andCh'i ch'iao t'u chieh (Tangram
solutions).Two volumes of tangrams and
solutions with no title page, Chinese labels of the puzzles, in Chinese format
(i.e. printed as long sheets on thin paper, accordion folded and stitched with
ribbon.Nd [c1820s??], stiff card
covers with flyleaves of a different paper, undoubtedly added later.84 pages in each volume, containing 334
problems and solutions.With ownership
stamp of a cartouche enclosing EWSHING, probably a Mr. E. W. Shing.Slocum says this is a c1820s reprint of the
earliest Chinese tangram book which appeared in 1813 & 1815.This version omits the TP and opening
text.I have a photocopy of the opening
material from Slocum.The original
problem book had a preface bySang‑hsia
K'o,which was repeated in the solution
book with the same date.Includes all
the problems of Shichi-kou-zu Gappeki, qv.

Child.Girl's Own Book.1833: 85;1839: 72;1842: 156."Chinese Puzzles _ These consist of pieces of wood in the form of
squares, triangles, &c.The object
is to arrange them so as to form various mathematical figures."

Anon.Edo Chiekata (How to Learn It??) (In
Japanese).Jan 1837, 19pp, 306
problems.(Unclear if this uses the
Tangram pieces.)Reprinted in the same
booklet as Sei Sh_nagon, on pp. 37‑55.

Augustus De
Morgan.On the foundations of algebra,
No. 1.Transactions of the Cambridge
Philosophical Society 7 (1842) 287-300.??NX. On pp. 289, he says
"the well-known toy called the Chinese Puzzle, in which a prescribed
number of forms are given, and a large number of different arrangements, of
which the outlines only are drawn, are to be produced."

Boy's
Treasury.1844.Puzzles and paradoxes, no. 16: The Chinese
puzzle, pp. 426-427.Instructions seem
to intend the tangrams, but they give five shapes and say to make one copy of
some and two copies of the others.As I
read it, this is incorrect, though it is intended to be the tangrams.11problem shapes given, no answers.Most of the shapes occur in earlier tangram collections, particularly in
A New Invented Chinese Puzzle.This
item is reproduced, complete with the error, as:Magician's Own Book, 1857, prob. 49, pp. 289-290;Landells, Boy's Own Toy-Maker, 1858, pp.
139-140;Book of 500 Puzzles, 1859, pp.
103-104;Boy's Own Conjuring Book,
1860, pp. 251-252;Wehman, New Book of
200 Puzzles, 1908, pp. 34-35.

Leske.Illustriertes Spielbuch für Mädchen.1864?

Prob.
584-11, pp. 288 & 405: Chinesisches Verwandlungsspiel.Make a square with the tangram pieces.Shows just five of the pieces, but correctly
states which two to make two copies of.

Prob.
584-18/25, pp. 289-291 & 407: Hieroglyphenspiele.Form various figures from various sets of pieces, mostly
tangrams, but the given shapes have bits of writing on them so the assembled
figure gives a word.Only one of the
shapes is as in Boy's Treasury.

Prob.
588, pp. 298 & 410: Etliche Knackmandeln.Another tangram problem like the preceeding, not equal to any in Boy's
Treasury.

Adams &
Co., Boston.Advertisement in The
Holiday Journal of Parlor Plays and Pastimes, Fall 1868.Details?? _ xerox sent by Slocum.P. 6: Chinese Puzzle.The celebrated Puzzle with which a hundred
or more symmetrical forms can be made, with book showing the designs.Though not illustrated, this seems likely to
be the Tangrams _ ??

J. Murray
(editor of the OED).Two letters to H.
E. Dudeney (9 Jun 1910&1 Oct 1910).The first inquires about the word 'tangram', following on
Dudeney's mention of it in his "World's best puzzles" (op. cit. in
2).The second says that 'tan' has no
Chinese origin;is apparently mid 19C,
probably of American origin;and the
word 'tangram' first appears in Webster's Dictionary of 1864. Dudeney, AM, 1917, p. 44, excerpts these
letters.

F. T. Wang
& C.‑S. Hsiung.A theorem on
the tangram.AMM 49 (1942) 596‑599.Determine the 20 convex regions which 16
isoceles right triangles can form and hence the 13 ones which the Tangram
pieces can form.

Mitsumasa
Anno.Anno's Math Games.(Translation of: Hajimete deau sugaku no
ehon; Fufkuinkan Shoten, Tokyo, 1982.)Philomel Books, NY, 1987.Pp.
38-43 & 95-96 show a simplified5-piece tangram-like puzzle which I have not seen before.The pieces are: a square of side1;three isosceles right triangles of side1;a right trapezium with bases1and2,altitude1and slant sideÖ2.The trapezium can be viewed as putting
together the square with a triangle.19
problems are set, with solutions at the back.

At the
International Congress on Mathematical Education, Seville, 1996, the
Mathematical Association gave outThe3, 4, 5Tangram,a cut card tangram, but in a6 x
8rectangular shape, so that the medium
sized triangle was a3-4-5
triangle.I modified this in Nov 1999,
by stretching along a diagonal to form a rhombus with angles double the angles
of a 3-4-5 triangle, so that four of the triangles are similar to 3-4-5
triangles.Making the small triangles be
actually 3-4-5, all edges are integral.I made up 35 problems with these pieces.I later saw that Hans Wierzorke has mentioned this dissection in
CFF, but with no problems.I
distributed this as my present at the Fourth Gathering for Gardner, Feb 2000.

6.S.1.LOCULUS
OF ARCHIMEDES

See S&B
22.I recall there is some dispute as
to whether the basic diagram should be a square or a double square.

E. J.
Dijksterhuis.Archimedes.Munksgaard, Copenhagen, 1956;reprinted by Princeton Univ. Press, 1987.Pp. 408‑412 is the best discussion of
this topic and supplies most of the classical references.

Archimedes.Letter to Eratosthenes, c-250?.Greek palimpsest, c975, on MS no. 355, from
the Cloister of Saint Sabba (= Mar Saba), Jerusalem, then at Metochion of the
Holy Sepulchre, Constantinopole.[This
MS disappeared in the confusion in Asia Minor in the 1920s but reappeared in
1998 when it was auctioned by Christie's in New York for c2M$.Hopefully, modern technology will allow a
facsimile and an improved transcription in the near future.]Described by J. L. Heiberg (& H. G.
Zeuthen); Eine neue Schrift des Archimedes; Bibliotheca Math. (3) 7 (1906‑1907)
321‑322.Heiberg describes the
MS, but only mentions the loculus.The
text is in Heiberg's edition of Archimedes; Opera; 2nd ed., Teubner, Leipzig,
1913, vol. II, pp. 415‑424, where it has been restored using the Suter
MSS below.Heath only mentions the
problem in passing.Heiberg quotes
Marius Victorinus, Atilius Fortunatianus and cites Ausonius and Ennodius.

H.
Suter.Der Loculus Archimedius oder das
Syntemachion des Archimedes.Zeitschr.
für Math. u. Phys. 44 (1899) Supplement= AGM 9 (1899) 491‑499.This is a collation from two 17C Arabic MSS which describe the
construction of the loculus. They are
different than the above MS.The German
is included in Archimedes Opera II, 2nd ed., 1913, pp. 420‑424.

Dijksterhuis
discusses both of the above and says that they are insufficient to determine
what was intended.The Greek seems to
indicate that Archimedes was studying the mathematics of a known puzzle.The Arabic shows the construction by cutting
a square, but the rest of the text doesn't say much.

Collins.Book of Puzzles.1927.The loculus of
Archimedes, pp. 7-11.Pieces made from
a double square.

6.S.2.OTHER
SETS OF PIECES

See Hoffmann & S&B, cited at
the beginning of 6.S, for general surveys.

See Bailey in 6.AS.1 for an 1858
puzzle with 10 pieces and The Sociable and Book of 500 Puzzles, prob. 10, in
6.AS.1 for an 11 piece puzzle.

There are many versions of this idea
available and some are occasionally given in JRM.

The Richter Anchor Stone puzzles and
building blocks were inspired by Friedrich FROEBEL (or Fröbel) (1782‑1852),
the educational innovator.He was the
inventor of Kindergartens, advocated children's play blocks and inspired both
the Richter Anchor Stone Puzzles and Milton Bradley.The stone material was invented by Otto Lilienthal (1848‑1896)
(possibly with his brother Gustav) better known as an aviation pioneer _ they
sold the patent and their machines to F. Adolph RICHTER for 1000 marks.The material might better be described as a
kind of fine brick which could be precisely moulded.Richter improved the stone and began production at Rudolstadt,
Thüringen, in 1882; the plant closed in 1964.Anchor was the company's trademark.He made at least 36 puzzles and perhaps a dozen sets of building blocks
which were popular with children, architects, engineers, etc.The Deutsches Museum in Munich has a whole
room devoted to various types of building blocks and materials, including the
Anchor blocks.There is an Anker Museum
in the Netherlands (Stichting Ankerhaus (= Anker Museum); Opaalstraat 2‑4
(or Postf. 1061), NL-2400 BB Alphen aan den Rijn, NETHERLANDS; tel: 01720‑41188)
which produces replacement parts for Anker stone puzzles.Modern facsimiles of the building sets are
also being produced.

In 1996 I
noticed the ceiling of the room to the south of the Salon of the Ambassadors in
the Alcazar of Seville.This 15C? ceiling
was built by workmen influenced by the Moorish tradition and has 112 square
wooden panels in a wide variety of rectilineal patterns.One panel has some diagonal lines and looks
like it could be used as a 10 piece tangram-like puzzle.Consider a4 x 4square.Draw both diagonal lines, then at two
adjacent corners, draw two lines making a unit square at these corners.At the other two corners draw one of these
two lines, namely the one perpendicular to their common side.This gives six isosceles right triangles of
edge1;two pentagons with three right angles and sides1, 2, 1, Ö2, Ö2;two quadrilaterals with two right angles and sides2, 1, Ö2, 2Ö2.Since geometric patterns and paneling are
common features of Arabic art, I wonder if there are any instances of such
patterns being used for a tangram-like puzzle?

Jackson.Rational Amusement.1821.Geometrical Puzzles, nos. 20-27, pp. 27-29 & 88-89 & plate II,
figs. 15-22.This is a set of 20 pieces
of 8 shapes used to make a square, a right triangle, three squares, etc.

Crambrook.1843.P. 4, no. 1: Pythagorean Puzzle, with Book.Though not illustrated, this is probably(??) the puzzle described
in Hoffmann, below, which was a Richter Anchor puzzle No. 12 of the same name
and is still occasionally seen.See
S&B 28.

Edward
Hordern has a Circassian Puzzle, c1870, with many pieces.

Hoffmann.1893.Chap. III, no. 3: The Pythagoras Puzzle, pp. 83-85 & 117-118.This has 7 pieces and is quite like the
Tangram _ see comment under Crambrook.

Loyd.Sam Loyd's Puzzle Magazine, January
1908.??NYS.(Given in A. C. White; Sam Loyd and His Chess Problems;
1913, op. cit. in 1; p. 100, where it is described as the only solution with 2
pieces in the 4 central squares.)

Ahrens, MUS
I 227, 1910, says he first had this in a letter from E. B. Escott
dated1 Apr 1909.(Moser, below, refers this to the 1st ed.,
1900, but this must be due to his not having seen it.)

C. H.
Bullivant.Home Fun, 1910, op. cit. in
5.S.Part VI, Chap. IV: No. 2: Another
draught puzzle, pp. 515 & 520.The
problem says "no three men shall be in a line, either horizontally or
perpendicularly".The solution
says "no three are in a line in any direction" and the diagram shows
this is indeed true.

Blyth.Match-Stick Magic.1921.Matchstick board
game, p. 73.6 x 6version phrased as putting "only two in
any one line: horizontal, perpendicular, or diagonal."However, his symmetric solution has three in
a row on lines of slope 2.

King.Best 100.1927.No. 69, pp. 28 &
55.Problem on the6 x 6board _ gives a symmetric solution.Says "there are two coins on every row" including
"diagonally across it", but he has three in a row on lines of slope
2.

Adams.Puzzle Book.1939.Prob. C.83: Stars
in their courses, pp. 144 & 181.Same solution as King, but he says "two stars in each vertical row,
two in each horizontal row, and two in each of the the two diagonals ....There must not be more than two stars in the
same straight line", but he has three in a row on lines of slope 2.

David R.
Nelson.Quasicrystals.SA 255:2 (Aug 1986) 32‑41 &
112.Exposits the discovery of
quasicrystals.First form is now called
'Shechtmanite'.

Kimberly-Clark
Corporation has taken out two patents on the use of the Penrose pattern for
quilted toilet paper as the non-repetition prevents the tissue from 'nesting'
on the roll.In Apr 1997, Penrose
issued a writ against Kimberly Clark Ltd. asserting his copyright on the
pattern and demanding damages, etc.

D. A.
Klarner.Brick‑packing
puzzles.JRM 6 (1973) 112‑117.General survey.In this he mentions a result that I gave him _ that2 x 3 x 7fills a8 x 11 x 21,but that the box cannot be divided into two
packable boxes.However, I gave him the
case1 x 3 x 4in5
x 5 x 12which is the smallest example
of this type.

6.V.SILHOUETTE
AND VIEWING PUZZLES

Viewing problems must be common among
draughtsmen and engineers, but I haven't seen many examples.I'd be pleased to see further examples.

2
silhouettes.

Circle&triangle_van Etten,Ozanam,Guyot,Magician's Own Book (UK version)

Circle&square_van Etten

Circle&rhombus_van Etten,Ozanam

Rectangle
with inner rectangle&rectangle with notch_Diagram Group.

3
silhouettes.

Circle,circle,circle_Madachy

Circle,cross,square_Wyatt,Perelman

Circle,oval,rectangle_van Etten,Ozanam,Guyot,Magician's Own Book (UK version)

Prob.
22 (misnumbered 15 in 1626) (Prob. 20), pp. 19‑20 & figs. opp.
p. 16 (pp. 35‑36): 2 silhouettes _one circular, the other triangular, rhomboidal or square.(English ed. omits last case.)The 1630 Examen says the author could have
done better and suggests:isosceles
triangle, several scalene triangles, oval or circle, which he says can be done
with an elliptically cut cone and a scalene cone.I am not sure I believe these.It seems that the authors are allowing the object to fill the hole and
to pass through the hole moving at an angle to the board rather than
perpendicularly as usually understood.In the English edition the Examination is combined with that of the next
problem.

Dudeney.Great puzzle crazes.Op. cit. in 2.1904.He sayssquare, circle and triangleis in a book in front of him dated
1674.I suspect this must be the 1674
English edition of van Etten, but I don't find the problem in the English
editions I have examined.Perhaps
Dudeney just meant that the idea was given in the 1674 book, though he is
specifically referring to the square, circle, triangle version.

Ozanam.1725.Vol. II, prob. 58 & 59, pp. 455‑458 & plate 25* (53 (note
there is a second plate with the same number)).Circle and triangle;circle and rhombus;circle,
oval, rectangle;circle, oval,
square.Figures are very like van
Etten.See Ozanam-Montucla, 1778.

1801.Item 536: Die 3 mathematischen Löcher.(See also the picture of Item 275, but that
text is for another item.)Square,
triangle and circle.

1807.Item 1126: Tricks includes thesquare, triangle and circle.

Badcock.Philosophical Recreations, or, Winter
Amusements.[1820].P. 14, no. 23: How to make a Peg that will
exactly fit three different kinds of Holes."Let one of the holes be circular, the other square, and the third
an oval; ...."Solution is a
cylinder whose height equals its diameter.

Jackson.Rational Amusement.1821.Geometrical Puzzles.

No.
16, pp. 26 & 86.Circle, square,
triangle,with discussion of the
dimensions:"a wedge, except that
its base must be a circle".

No.
29, pp. 30 & 89-90.Circle, oval,
square.

Endless
Amusement II.1826?P. 62:"To make a Peg that will exactly fit three different kinds of
Holes."Circle, oval, square.c= Badcock.

F. J. P.
Riecke.Op. cit. in 4.A.1, vol. 1,
1867.Art. 33: Die Ungula, pp. 58‑61.Take a cylinder with equal height and
diameter.A cut from the diameter of
one base which just touches the other base cuts off a piece called an ungula
(Latin for claw).He computes the
volume as4r3/3.He then makes the symmetric cut to produce
the circle, square, triangle shape, which thus has volume(2π ‑ 8/3) r3.Says he has seen such a shape and a board
with the three holes as a child's toy.Cf. Simpson, 1745.

Magician's
Own Book (UK version).1871.The round peg in the square hole:To pass a cylinder through three different
holes, yet to fill them entirely, pp. 49-50.Circle, oval, rectangle;circle
& (isosceles) triangle.

Alfred
Elliott.Within‑Doors.A Book of Games and Pastimes for the Drawing
Room.Nelson, 1872.[Toole Stott 251.Toole Stott 1030 is a 1873 ed.]No. 4: The cylinder puzzle, pp. 27‑28 & 30‑31.Circle, square, triangle.

Williams.Home Entertainments.1914.The plug puzzle, pp. 103-104.Circle, square, triangle and rectangle with curved ends.This is the only example of this four-fold
form that I have seen.Nice drawing of
a board with the plug shown in each hole, except the curve on the sloping faces
is not always drawn down to the bottom.

Ernest R.
Ranucci.Non‑unique orthographic
projections.RMM 14 (Jan‑Feb
1964) 50.3 views such that there
are 10 different objects with these views.

The Diagram
Group.The Family Book of Puzzles.The Leisure Circle Ltd., Wembley, Middlesex,
1984.Problem 114, with Solution at the
back of the book.Front view is a
rectangle with an interior rectangle.Side view is a rectangle with a rectangular notch on front side.Solution is a short cylinder with a straight
notch in it.This is a fairly classic
problem for engineers but I haven't seen it in print elsewhere.

Marek
Penszko.Polish your wits _ 3: Loop the
loop.Games 11:2 (Feb/Mar 1987) 28
& 58.Draw lines on a glass cube to
produce three given projections.Problem asks for all three projections to be the same.

6.W.BURR
PUZZLES

When assembled, a burr looks like
three sticks crossing orthogonally, forming a 'star' with six points at the
vertices of an octahedron.Slocum says
Wyatt [Puzzles in Wood, 1928, op. cit. in 5.H.1] is the first to use the word
'burr'.Collins, Book of Puzzles, 1927,
p. 135, calls them "Cluster, Parisian or Gordian Knot Puzzles" and
states: "it is believed that they were first made in Paris, if, indeed,
they were not invented thre."

See S&B, pp. 62‑85.

See also 6.BJ.

6.W.1.THREE
PIECE BURR

Most of these have three pieces which
are rectangular in cross-section with slots of the same size and some of the
pieces have notches from the slot to the outside.When one piece is pushed, it slides, revealing its notch.When placed properly, this allows a second
piece to slide off and out.In the
1990s, a more elaborate type of three piece burr has appeared.These have three3 x 3 x 5pieces which
intersect in a central3 x 3 x 3region.Within this region, some of the unit cubes are not present, which allows
sliding of the pieces.Some versions of
the puzzle permit twisting of pieces though this usually requires a bit of
rounding of edges and the actual examples tend to break, so these are not as acceptable.

Crambrook.1843.P. 5, no. 4:Puzzling Cross3 pieces.This seems likely to be a three piece burr, but perhaps is in 6.W.3 _
??It is followed by"Maltese Cross6 pieces".

Edward
Hordern has examples in ivory from 1850-1900.

Hoffmann.1893.Chap. III, no. 35: The cross‑keys or three‑piece puzzle, pp.
106 & 139.(Hordern, p. 67, has a
photo.)One piece has an extra small
notch which does not appear in other versions where the dimensions are better
chosen.

Benson.1904.The cross keys puzzle, pp. 205‑206.

Pearson.1907.Part III, no. 56: The cross‑keys, pp. 56 & 127‑128.

Arthur Mee's
Children's Encyclopedia 'Wonder Box'.The Children's Encyclopedia appeared in 1908, so this is probably 1908
or soon thereafter.3-Piece Mortise
with thin pieces.

Anon.Woodwork Joints.Evans, London, (1918), 2nd ed., 1919.[I have also seen a 4th ed., 1925, which is identical to the 2nd
ed., except for advertising pages at the end.]A mortising puzzle, pp. 197‑199.

Dic
Sonneveld seems to be the first to begin designing three piece burrs of the
more elaborate style, perhaps about 1985.Trevor Wood has made several examples for sale.

Bill
Cutler.Email announcement to NOBNET on
27 Jan 1999.He has begun analysing the
newer style of three piece burr, excluding twist moves.His first stage has examined cases where the
centre cube of the central region is occupied and the piece this central cube
belongs to has no symmetry.He
finds202 x 109assemblies (I'm not sure if this is an exact
figure) and there are33level-8 examples (i.e. where it takes8moves to remove the first piece);6674level-7 examples;73362level-6 examples.He thinks this
is about70%of the total and it is already about six times the number of
cases considered for the six piece burr (see 6.W.2).

Bill
Cutler.Christmas letter of 4 Dec
1999.Says he has completed the above
analysis and found25 x 1010possibilities, which took 225 days on a
workstation.The most elaborate
examples require 8 moves to get a piece out and there are 80 of these.He used one for his IPP19 puzzle.

Catel.Kunst-Cabinet.1790.Die kleine
Teufelsklaue, p. 10 & fig. 16 on plate I.Figure shows it assembled and fails to draw one of the divisions between
pieces.Description says it is 6
pieces, 2 inches long, from plum wood and costs 3 groschen (worth about an
English penny of the time).(See also
pp. 9-10, fig. 20 on plate I for Die grosse Teufelsklaue _ the 'squirrelcage'.)

Bestelmeier.1801.Item 147: Die kleine Teufelsklaue.(Note _ there is another item 147 on the next plate.)Only shows it assembled.Brief text may be copying part of
Catel.See also the picture for item
1099 which looks like a six‑piece burr included in a set of puzzles.(See also Item 142: Die grosse
Teufelsklaue.)

Edward
Hordern has examples, called The Oak of Old England, from c1840.

Crambrook.1843.P. 5, no. 5:Maltese Cross6 [pieces], three sorts.Not clear if these might be here or in 6.W.4
or 6.W.5 _ ??

Magician's
Own Book.1857.Prob. 1: The Chinese cross, pp. 266-267
& 291.One plain key piece.Not the same as in Minguét.

Landells.Boy's Own Toy-Maker.1858.Pp. 137-139.Identical to
Magician's Own Book.

Book of 500
Puzzles.1859.1: The Chinese cross, pp. 80-81 &
105.Identical to Magician's Own Book.

A. F.
Bogesen (1792‑1876).In the
Danish Technical Museum, Helsingør (= Elsinore) are a number of wooden puzzles
made by him, including a 6 piece burr, a 12 piece burr, an Imperial Scale? and
a complex (trick??) joint.

Dudeney.Prob. 473 _ Chinese cross.Weekly Dispatch (23 Nov&7 Dec 1902), both p. 13.Different than one known to his correspondents.

Dudeney.Great puzzle crazes.Op. cit. in 2.1904."... the
"Chinese Cross," a puzzle of undoubted Oriental origin that was
formerly brought from China by travellers as a curiosity, but for a long time
has had a steady sale in this country."

William H.
[Bill] Cutler.The six‑piece
burr.JRM 10 (1977‑78) 241‑250.Complete, computer assisted, analysis, with
help from T. H. O'Beirne and A. C. Cross.Pieces are considered as 'notchable' if they can be made by a sequence
of notches, which are produced by two saw cuts and then chiseling out the space
between them.Otherwise viewed, notches
are what could be produced by a wide cutter or router.There are25of these which can occur in
solutions.(In 1994, he states that
there are a total of59notchable pieces and diagrams all of
them.)One can also have more general
pieces with 'right-angle notches' which would require four chisel cuts _ e.g.
to cut a single1 x 1 x
1piece out of a2 x 2 x 8rod.Alternatively, one can glue
cubes into notches.There are369which can occur in solutions.(In 1994, he states that there are837pieces which produce2225different oriented pieces, and he lists them all.)He only considers solid solutions _ i.e.
ones where there are no internal holes.He finds and lists the314'notchable' solutions.There are119,979general solutions.

C. Arthur
Cross.The Chinese Cross.Pentangle, Over Wallop, Hants., UK,
1979.Brief description of the
solutions in the general case, as found by Cutler and Cross.

Bill
Cutler.Holey 6‑Piece Burr!Published by the author, Palatine,
Illinois.(1986);with addendum, 1988, 48pp.He is now permitting internal holes.Describes holey burrs with notchable pieces,
particularly those with multiple moves to release the first piece.

Bill
Cutler.A Computer Analysis of All
6-Piece Burrs.Published by the author,
ibid., 1994.86pp.Sketches complete history of the
project.(I have included a few details
in the description of his 1977/78 article, above.)In 1987, he computed all the notchable holey solutions, using
about 2 months of PC AT time, finding13,354,991assemblies giving7.4 million solutions.Two of these were level 10 _ i.e. they
require 10 moves to remove the first piece (or pieces), but the highest level
occurring for a unique solution was 5.After that he started on the general holey burrs and estimated it would
take 400 years of PC AT time _ running at 8 MHz.After some development, the actual time used was about 62.5 PC AT
years, but a lot of this was done on by Harry L. Nelson during idle time on the
Crays at Lawrence Livermore Laboratories, and faster PCs became available, so
the whole project only took about 2½ years, being completed in Aug 1990 and
finding35,657,131,235assemblies.He hasn't checked if all assemblies come apart fully, but he estimates
there are 5.75 billion solutions.He
estimates the project used 45 times the computing power used in the proof of
the Four Color Theorem and that the project would only take two weeks on the
eight RS6000 workstations he now supervises.Some 70,000 high-level solutions were specifically saved and can be obtained
on disc from him.The highest level
found was 12 and the highest level for a unique solution was 10.See 6.W.1 for a continuation of this work.

6.W.3.THREE
PIECE BURR WITH IDENTICAL PIECES

See S&B, p. 66.

Crambrook.1843.P. 5, no. 4:Puzzling Cross3 pieces.This seems likely to be a three piece burr, but perhaps is in 6.W.1 _
??It is followed by"Maltese Cross6 pieces".

Wilhelm
Segerblom.Trick wood joining.SA (1 Apr 1899) 196.

6.W.4.DIAGONAL
SIX PIECE BURR=TRICK STAR

This version often looks like a
stellated rhombic dodecahedron.It has
two basic forms, one with a key piece;the other with all pieces identical, which assembles as two groups of
three.

See S&B, p. 78.

Crambrook.1843.P. 5, no. 5:Maltese Cross6 [pieces], three sorts.Not clear if these belong here or in 6.W.2
or 6.W.5 _ ??

Slocum.Compendium.Shows Star Puzzle from The Youth's Companion, 1875.The picture does not show which form it is.

S. P.
Chandler.US Patent 393,816 _
Puzzle.Patented 23 Apr 1888.1p + 1p diagrams, but the text page is
missing from my copy _ get??.Coffin
says this is the earliest version, but it is more complex than usual, with12pieces, and has a key piece.

Iffland
Frères (Lausanne).Swiss patent 245,402
_ Zusammensetzspiel.Received
19 Nov 1945;granted 15 Nov
1946;published 1 Jul 1947.2pp + 1p diagrams.Stellated rhombic dodecahedral version with a key piece.(Coffin says this is the first to use this
shape, although Slocum has a version c1875.)

6.W.5.SIX
PIECE BURR WITH IDENTICAL PIECES

One form has six identical pieces and
all move outward or inward together.Another form with flat notched pieces has one piece with an extra notch
or an extended notch which allows it to fit in last, either by sliding or
twisting, but this is not initially obvious.This form is sometimes made with equal pieces so that it can only be
assembled by force, perhaps after steaming, and it then makes an unopenable
money box.This might be considered
under 11.M.

Edward
Hordern has a version with one piece a little smaller than the rest from c1800.

Crambrook.1843.P. 5, no. 5:Maltese Cross6 [pieces], three sorts.Not clear if these belong here or in 6.W.2
or 6.W.4 _ ??

C.
Baudenbecher catalogue, c1850s.Op.
cit. in 6.W.7.This has an example of
the six equal flat pieces making an unopenable(?) money box.

Bestelmeier.1801.Item 142: Die grosse Teufelsklaue.The 'squirrelcage', identical to Catel, with same drawing, but reversed.Text may be copying some of Catel.

C.
Baudenbecher, toy manufacturer in Nuremberg.Sample book or catalogue from c1850s.Baudenbecher was taken over by J. W. Spear & Sons in 1919 and the
catalogue is now in the Spear's Game Archive, Ware, Hertfordshire.It comprises folio and double folio sheets
with finely painted illustrations of the firm's products.One whole folio page shows about 20 types of
wooden interlocking puzzles, including most of the types mentioned elsewhere in
this section and in 6.W.5 and 6.BJ.Until I get a picture, I can't be more specific.

Slocum.Compendium.Shows a 'woodchuck' type puzzle, called White Wood Block Puzzle, from
The Youth's Companion, 1875.I can't
see how many pieces it has:12 or 18??

Sidney
Melmore.A single‑sided doubly
collapsible tessellation.MG 31 (No.
294) (1947) 106.Forms a Möbius strip
of three triangles and three rhombi, which is basically a flexagon (cf 6.D).He sees it has two distinct forms, but
doesn't see the flexing property!!He
describes how to extend these hexagons into a tessellation which has some
resemblance to other items in this section.

Wallace G.
Walker invented his "IsoAxis" ® in 1958 while a student at Cranbrook
Academy of Art, Michigan.This is
approximately a ring of ten tetrahedra.He obtained a US Patent for it in 1967 _ see below.In 1973(?) he sent an example to Doris
Schattschneider who soon realised that the basic idea was a ring of tetrahedra
and that Escher tessellations could be adapted to it.They developed the idea into "M. C. Escher
Kaleidocycles", published by Ballantine in 1977 and reprinted several
times since.

Douglas
Engel.Flexahedrons.RMM 11 (Oct 1962) 3‑5.These have 'Jacob's ladder' hinges, not edge‑to‑edge
hinges.He says he invented these in
Fall, 1961.He formed rings of4, 6, 7, 8tetrahedra and used a diagonal joining to make rings of 4 and 6 cubes.

Frederick
George Flowerday.US Patent 3,916,559 _
Vortex Linkages.Filed: 12 Aug 1974 (23
Aug 1973 in UK);issued: 4 Nov
1975.Abstract + 2pp + 3pp
diagrams.Mostly shows his Hexyflex,
essentially a six piece ring of tetrahedra, but with just four edges of each
tetrahedron present.He also shows his
Octyflex which has eight pieces.Text
refers to any even number³ 6.

Pacioli.Summa.1494.Part 2, f. 55r, prob.
33.Florence is 5 miles around the inside.The wall is3½braccia wide and the ditch
is14braccia wide _ how far is it around the outside?Several other similar problems.

"A
Lover of the Mathematics."A
Mathematical Miscellany in Four Parts.2nd ed., S. Fuller, Dublin, 1735.The First Part is:An Essay
towards the Probable Solution of the Forty five Surprising PARADOXES, in
GORDON's Geography, so the following must have appeared in Gordon.Part I, no. 73, p. 56."'Tis certainly Matter of Fact, that
three certain Travellers went a Journey, in which, Tho' their Heads travelled full
twelve Yards more than their Feet, yet they all return'd alive, with their
Heads on."

Carlile.Collection.1793.Prob. XXV, p. 17.Two men travel, one upright, the other
standing on his head.Who "sails
farthest"?Basically he compares
the distance travelled by the head and the feet of the first man.He notes that this argument also applies to
a horse working a mill by walking in a circle; the outside of the horse travels
about six times the thickness of the horse further than the inside on each
turn.

Jackson.Rational Amusement.1821.Geographical Paradoxes, no. 54, pp. 46 & 115-116."It is a matter of fact, that three
certain travellers went on a journey, in which their heads travelled full
twelve yards more than their feet; and yet, they all returned alive with their
heads on."Solution says this is
discussed in Whiston's Euclid, Book 3, Prop. 37, Schol. (3.).[This first appeared in 1702.]

K. S.
Viwanatha Sastri.Reminiscences of my
esteemed tutor.In:P. K. Srinivasan, ed.; Ramanujan Memorial
Volumes:1: Ramanujan _ Letters and
Reminiscences;2: Ramanujan _ An
Inspiration;Muthialpet High School,
Number Friends Society, Old Boys' Committee, Madras, 1968.Vol. 1, pp. 89-93.On p. 93, he relates that this was a favourite problem of his
tutor, Srinivasan Ramanujan.Though not
clearly dated, this seems likely to be c1908-1910, but may have been up to
1914."Suppose we prepare a belt
round the equator of the earth, the belt being2πfeet longer, and if we
put the belt round the earth, how high will it stand?The belt will stand1foot high, a substantial
height."

Ludwig
Wittgenstein was fascinated by the problem and used to pose it to
students.Most students felt that
adding a yard to the rope would raise it from the earth by a negligible amount
_ which it is, in relation to the size of the earth, but not in relation to the
yard.See:John Lenihan;Science
in Focus;Blackie, 1975, p. 39.

Ernest K.
Chapin.Loc. cit. in 5.D.1.1927.Prob. 5, p. 87 & Answers p. 7.A yard is added to a band around the earth.Can you raise it 5 inches?Answer notes the size of the earth is immaterial.

Collins.Book of Puzzles.1927.The globetrotter's
puzzle, pp. 68‑69.If you walk
around the equator, how much farther does your head go?

Prob.
20: A global readjustment.Take a wire
around the earth and insert an extra 40 ft into it _ how high up will it be?

Prob.
23: Getting ahead.If you walk around
the earth, how much further does your head go than your feet?

W. A.
Bagley.Puzzle Pie.Op. cit. in 5.D.5.1944.Things are seldom
what they seem _ No. 42a, 43, 44, pp. 50-51.42a and 43 ask how much the radius increases for a yard gain of
circumference.No. 44 asks if we add a
yard to a rope around the earth and then tauten it by pulling outward at one
point, how far will that point be above the earth's surface?

Birtwistle.Math. Puzzles & Perplexities.1971.Find the angle, pp. 86-87.Short
solution using law of sines and other simple trigonometric relations.

Colin
Tripp.Adventitious angles.MG 59 (No. 408) (Jun 1975) 98‑106.Studies when Ð CEDcan be determined and all angles are an
integral number of degrees.Computer
search indicates that there are at most53cases.

D. A. Q.
[Douglas A. Quadling].The adventitious
angles problem: a progress report.MG
61 (No. 415) (Mar 1977) 55-58.Reports
on a number of contributions resolving the cases which Tripp could not
prove.All the work is complicated
trigonometry _ no further cases have been demonstrated geometrically.

J[ohn]. F.
Rigby.Adventitious quadrangles: a
geometrical approach.MG 62 (No. 421)
(Oct 1978) 183-191.Gives geometrical
proofs for almost all cases.Cites Bol
and a long paper of his own to appear in Geom. Dedicata (??NYS).He drops the condition thatABCbe isosceles.His adventitious
quadrangles correspond to Bol's triple intersections of diagonals of a
regularn-gon.

MS 27:3
(1994/5) 65has two straightforward
letters on the problem, which was mentioned in ibid. 27:1 (1994/5) 7.One letter cites 1938 and 1955 appearances.P. 66 gives another solution of the
problem.See next item.

Douglas
Quadling.Letter: Langley's
adventitious angles.MS 27:3 (1994/5)
65‑66.He was editor of MG when
Tripp's article appeared.He gives some
history of the problem and some life of Langley (d. 1933).Edward Langley was a teacher at Bedford
Modern School and the founding editor of the MG in 1894-1895.E. T. Bell was a student of Langley's and
contributed an obituary in the MG (Oct 1933) saying that Langley was the finest
expositor he ever heard _ ??NYS.Langley also had botanical interests and a blackberry variety is named
for him.

Albrecht
Dürer.Elementorum Geometricorum (?) _
the copy of this that I saw at the Turner Collection, Keele, has the title page
missing, but Elementorum Geometricorum is the heading of the first text page
and appears to be the book's title.This appears to be a Latin translation of Unterweysung der Messung
....Christianus Wechelus, Paris,
1532.Liber quartus, fig. 29-43, pp.
145-158 shows the same material as in the 1525 edition.

Cardan.De Rerum Varietate.1557, ??NYS= Opera Omnia, vol. III, pp. 246-247.Liber XIII.Corpora, qua
regularia diei solent, quomodo in plano formentur.Shows nets of the regular solids, except the two halves of the
dodecahedron have been separated to fit into one column of the text.

Pike.Arithmetic.1788.Pp. 458-459."As the figures of some of these bodies
would give but a confused idea of them, I have omitted them; but the following
figures, cut out in pasteboard, and the lines cut half through, will fold up
into the several bodies."Gives
the regular polyhedra.

Perelman.FMP.c1935?To develop a cube, pp.
179 & 182‑183.Asserts there
are10nets and draws them, but two "can be turned upside down and this
will add two more ...."One shape
is missing.Of the two marked as
reversible, one is symmetric, hence equal to its reverse, but the other isn't.

M.
Gardner.SA (Nov 1966) c= Carnival, pp.
41‑54.Discusses the nets of the
cube and the Answers show all11of them.He asks what shapes these11hexominoes will form _ they cannot form any
rectangles.He poses the four
dimensional problem;the Addendum says
he got several answers, no two agreeing.

A. Sanders
& D. V. Smith.Nets of the
octahedron and the cube. MTg 42 (Spring 1968) 60‑63.Finds11nets for the octahedron and
shows a duality with the cube.

Peter
Turney.Unfolding the tesseract.JRM 17 (1984‑85) 1‑16.Finds261nets of the4‑cube.(I don't believe this has ever been confirmed.)

P. Light
& D. Singmaster.The nets of the
regular polyhedra.Presented at New
York Acad. Sci. Graph Theory Day X, 213 Nov 1985.In Notes from New York Graph Theory Day X, 23 Nov 1985;
ed. by J. W. Kennedy & L. V. Quintas; New York Acad. Sci., 1986,
p. 26.Based on Light's BSc
project in 1984-1984 under my supervision.Shows there are43,380nets for the dodecahedron and
icosahedron.I may organize this into a
paper, but several others have since verified the result.

6.AB.SELF‑RISING
POLYHEDRA

H.
Steinhaus.Mathematical Snapshots.Stechert, NY, 1938.(= Kalejdoskop Matematyczny.Ksi__nica‑Atlas, Lwów and Warsaw,
1938, ??NX.)Pp. 74-75 describes the
dodecahedron and says to see the model in the pocket at the end, but makes no
special observation of the self-rising property.Described in detail with photographs in OUP, NY, eds:1950: pp. 161-164;1960: pp. 209‑212;1969 (1983): pp. 196-198.

M. Kac.Hugo Steinhaus _ a reminiscence and a
tribute.AMM 81 (1974) 572‑581.Material is on pp. 580‑581, with
picture on p. 581.

A pop‑up
octahedron was used by Waddington's as an advertising insert in a trade journal
at the London Toy Fair about 1981.Pop-up cubes have also been used.

6.AC.CONWAY'S
LIFE

There is now a web page devoted to
Life run by Bob Wainwright _ address is:

http://members.aol.com/life1ine/life/lifepage.htm
[sic!].

M.
Gardner.Solitaire game of
"Life".SA (Oct 1970).On cellular automata, self‑reproduction,
the Garden‑of‑Eden and the game of "Life".SA (Feb 1971).c= Wheels, chap. 20-22.In the Oct 1970 issue, Conway offered a $50 prize for a configuration
which became infinitely large _ Bill Gosper found the glider gun a month
later.At the Second Gathering for
Gardner, Atlanta, 1996, Bob Wainwright showed a picture of Gosper's telegram to
Garnder on 4 Nov 1970 giving the coordinates of the glider gun.I wasn't clear if Wainwright has this or
Gardner still has it.

Bays has
started a quarterly 3‑D Life Newsletter, but I have only seen one (or
two?) issues.??get??

Alan
Parr.It's Life _ but not as we know
it.MiS 21:3 (May 1992) 12-15.Life on a hexagonal lattice.

6.AD.ISOPERIMETRIC
PROBLEMS

There is quite a bit of classical
history which I have not yet entered.Magician's Own Book notes there is a connection between the Dido version
of the probelm and Cutting a card so one can pass through it, Section. 6.BA.

Virgil.Aeneid.‑19.Book 1, lines 360‑370.(p. 38 of the Penguin edition, translated by
W. F. Jackson Knight, 1956.)Dido
came to a spot in Tunisia and the local chiefs promised her as much land as she
could enclose in the hide of a bull.She cut it into a long strip and used it to cut off a peninsula and
founded Carthage.This story was later
adapted to other city foundations.John
Timbs; Curiosities of History; With New Lights; David Bogue, London, 1857,
devotes a section to Artifice of the thong in founding cities, pp. 49-50,
relating that in 1100, Hengist, the first Saxon King of Kent, similarly
purchased a site called Castle of the Thong and gives references to Indian,
Persian and American versions of the story as well as several other English
versions.

Pappus.c290.Synagoge [Collection].Book V,
Preface, para. 1‑3, on the sagacity of bees.Greek and English in SIHGM, Vol. II, pp. 588‑593.A different, abridged, English version is in
HGM II 389‑390.

The 5C Saxon
mercenary, Hengist or Hengest, is said to have requested from Vortigern:
"as much land as can be encircled by a thong".He "then took the hide of a bull and cut
it into a single leather thong.With
this thong he marked out a certain precipitous site, which he had chosen with
the greatest possible cunning."This is reported by Geoffrey of Monmouth in the 12C and this is quoted
by the editor in:The Exeter Book
Riddles;8-10C (Bryant (op cit in 9.E)
gives last quarter of the 10C) _ ??;Translated and edited by Kevin Crossley-Holland;(As: The Exeter Riddle Book, Folio Society,
1978, Penguin, 1979);Revised ed.,
Penguin, 1993; pp. 101-102.

Buteo.Logistica.1559.Prob. 86, pp.
298-299.If 9 pieces of wood are
bundled up by5½feet of cord, how much cord is needed to
bundle up 4 pieces?5 pieces?

Pitiscus.Trigonometria.Revised ed., 1600, p. 223.??NYS _ described in: Nobuo Miura; The applications of trigonometry in
Pitiscus: a preliminary essay; Historia Scientarum 30 (1986) 63-78.A square of side4and triangle of
sides5, 5, 3have the same perimeter but different areas.Presumably he was warning people not to be
cheated in this way.

Question
1, 1725: 327.Question 3, 1778:
328;1803: 325;1814: 276; 1840: 141.String twice as
long contains four times as much asparagus.

Question
2, 1725: 328.If a cord of length 10
encloses 200, how much does a cord of length 8 enclose?

Question
3, 1725: 328.Sack 5 high by 4 across
versus 4 sacks 5 high by 1 across.c= Q. 2, 1778: 328;1803: 324;1814: 276;1840: 140-141, which has sack 4 high by 6
around versus two sacks 4 high by 3 around.

Question
4, 1725: 328‑329.How much water
does a pipe of twice the diameter deliver?

Les
Amusemens.1749.

Prob.
211, p. 376.String twice as long
contains four times as much asparagus.

Prob.
212, p. 377.Determine length of string
which contains twice as much asparagus.

Prob.
223-226, pp. 386-389.Various problems
involving changing shape with the same perimeter.Notes the area can be infinitely small.

No.
30, pp. 30 & 90.Square field versus
oblong (rectangular?) field of the same perimeter.

No.
31, pp. 30 & 90-91.String twice as
long contains four times as much asparagus.

Magician's
Own Book (UK version).1871.To cut a card for one to jump through, p.
124, says:"The adventurer of old,
who, inducing the aborigines to give him as much land as a bull's hide would
cover, and made it into one strip by which acres were enclosed, had probably
played at this game in his youth."See 6.BA.

6.AD.1.LARGEST PARCEL ONE CAN POST

New section.Are there older examples?

Richard A.
Proctor.Greatest content with parcels'
post.Knowledge 3 (3 Aug 1883) 76.Height + girth£6 ft.States that a cylinder is well known to be the best solution.Either for a cylinder or a box, the optimum
hasheight = 2,girth = 4,with optimum volumes2and8/π = 2.54... ft3.

H. F.Letter:Parcel post problem.Knowledge 3
(24 Aug 1883) 126, item 905.Suppose
'length' means "the maximum distance in a straight line between any two
points on its surface".By this he
means the diameter of the solid.Then
the optimum shape is the intersection of a right circular cylinder with a
sphere, the axis of the cylinder passing through the centre of the sphere, and
this has the 'length' being the diameter of the sphere and the maximum volume
is then2_ ft3

Algernon
Bray.Letter:Greatest content of a parcel which can be sent by post.Knowledge 3 (7 Sep 1883) 159, item 923.Says the problem is easily solved without
calculus.However, for the box, he says
"it is plain that the bulk of half the parcel will be greatest when [its]
dimensions are equal".

I have
looked at the current parcel post regulations and they saylength £ 1.5mandlength + girth £
3m,for which the largest box is1 x ½ x ½,with volume1/4.The largest cylinder has length1and
radius1/πwith volume1/π.

I have also
considered the simple question of a person posting a fishing rod longer than
the maximal length by putting it diagonally in a box.The longest rod occurs at a boundary maximum, at3/2 x 3/4 x 0or3/2 x 0 x 3/4,so one can post a rod of length3Ö5/4 = 1.677...,
which is about12%longer than1.5m.In this problem, the use
of a cylinder actually does worse!

6.AE.6"HOLE THROUGH SPHERE LEAVES CONSTANT VOLUME

Hamnet
Holditch.Geometrical theorem.Quarterly J. of Pure and Applied Math. 2
(1858) ??NYS, described by Broman.If a
chord of a closed curve, of constant lengtha+b,be divided into two parts
of lengthsa, brespectively, the difference between the
areas of the closed curve, and of the locus of the dividing point as the chord
moves around the curve, will beπab.When the closed curve
is a circle anda = b,then this is the two dimensional version
given by Jones, below.A letter from
Broman says he has found Holditch's theorem cited in 1888, 1906, 1975 and 1976.

Samuel I.
Jones.Mathematical Nuts.1932.P. 86.??NYS.Cited by Gardner, (SA, Nov 1957) = 1st
Book, chap. 12, prob. 7.Gardner says
Jones, p. 93, also gives the two dimensional version:If the longest line that can be drawn in an annulus is6"long, what is the area of the annulus?

L.
Lines.Solid Geometry.Macmillan, London, 1935;Dover, 1965.P. 101, Example 8W3:"A napkin ring is in the form of a sphere pierced by a cylindrical
hole.Prove that its volume is the same
as that of a sphere with diameter equal to the length of the hole."Solution is given, but there is no
indication that it is new or recent.

L. A.
Graham.Ingenious Mathematical Problems
and Methods.Dover, 1959.Prob. 34: Hole in a sphere, pp. 23 & 145‑147.[The material in this book appeared in
Graham's company magazine from about 1940, but no dates are provided in the
book.(??can date be found out.)]

C. W.
Trigg.Op. cit. in 5.Q.1967.Quickie 217: Hole in sphere, pp. 59 & 178‑179.Gives an argument based on surface tension
to see that the ring surface remains spherical as the hole changes radius.Problem has a 10" hole.

Tangential
chord, pp. 71-73.10"chord in an annulus.What is the area of the annulus?Does traditionally and then by letting inner
radius be zero.

The
hole in the sphere, pp. 87-88 & 177-178.Bore a hole through a sphere so the remaining piece has half the volume
of the sphere.The radius of the hole
is approx..61of the radius of the sphere.

Another
hole, pp. 89, 178 & 192.6"hole cut out of
sphere.What is the volume of the
remainder?Refers to the tangential
chord problem.

Arne
Broman.Holditch's theorem: An
introductory problem.Lecture at ICM,
Helsinki, Aug 1978.Broman then
sent out copies of his lecture notes and a supplementary letter on 30 Aug
1978.He discusses Holditch's proof
(see above) and more careful modern versions of it.His letter gives some other citations.

6.AF.WHAT
COLOUR WAS THE BEAR?

A hunter goes 100 mi south, 100 mi
east and 100 mi north and finds himself where he started.He then shoots a bear _ what colour was the
bear?

I include other polar problems
here.See also 10.K for related
geographical problems.

"A
Lover of the Mathematics."A
Mathematical Miscellany in Four Parts.2nd ed., S. Fuller, Dublin, 1735.The First Part is:An Essay
towards the Probable Solution of the Forty five Surprising PARADOXES, in
GORDON's Geography, so the following must have appeared in Gordon.Part I, no. 10, p. 9."There is a particular Place of the
Earth where the Winds (tho' frequently veering round the Compas) do always blow
from the North Point."

Philip
Breslaw (attrib.).Breslaw's Last Legacy;
or the Magical Companion: containing all that is Curious, Pleasing,
Entertaining and Comical; selected From the most celebrated Masters of
Deception: As well with Slight of Hand, As with Mathematical Inventions.Wherein is displayed The Mode and Manner of
deceiving the Eye; as practised by those celebrated Masters of Mirthful
Deceptions.Including the various
Exhibitions of those wonderful Artists, Breslaw, Sieur, Comus, Jonas, &c.Also the Interpretation of Dreams,
Signifcation of Moles, Palmestry, &c.The whole forming A Book of real Knowledge in the Art of
Conjuration.(T. Moore, London, 1784,
120pp.)With an accurate Description of
the Method how to make The Air Balloon, and inject the Inflammable Air.(2nd ed., T. Moore, London, 1784, 132pp;5th ed., W. Lane, London, 1791, 132pp.)A New Edition, with great Additions and
Improvements.(W. Lane, London, 1795,
144pp.)Facsimile from the copy in the
Byron Walker Collection, with added Introduction, etc., Stevens Magic Emporium,
Wichita, Kansas, 1997, HB.[This was
first published in 1784, after Breslaw's death, so it is unlikely that he had
anything to do with the book.There
were versions in1784, 1791, 1792,
1793, 1794, 1795, 1800, 1806, c1809, c1810, 1811, 1824.Hall, BCB 39-43, 46-51.Toole Stott 120-131, 966‑967.Heyl 35-41.This book went through many variations of subtitle and contents _ the
above is the largest version.].I will
cite the date as1784?.

Geographical Paradoxes.

Paradox
I, p. 35.Where is it noon every half
hour?Answer: At the North Pole in
Summer, when the sun is due south all day long, so it is noon every moment!

Paradox
II, p. 36.Where can the sun and the
full moon rise at the same time in the same direction?Answer: "Under the North Pole, the sun
and the full moon, both decreasing in south declination, may rise in the
equinoxial points at the same time; and under the North Pole, there is no other
point of compass but south."I
think this means at the North Pole at the equinox.

Carlile.Collection.1793.Prob. CXVI, p. 69.Where does the wind always blow from the
north?

Mr. X.His Pages.The Royal Magazine 10:3 (Jul 1903) 246-247.A safe catch.Airship
starts at the North Pole, goes south for seven days, then west for seven
days.Which way must it go to get back
to its starting point?No solution
given.

Pearson.1907.

Part
II, no. 21: By the compass, pp. 18 & 190.Start at North Pole and go20miles southwest.What direction gets back to the Pole the
quickest?Answer notes that it is hard
to go southwest from the Pole!

Part
II, no. 15: Ask "Where's the north?" _ Pope, pp. 117 & 194.Start1200miles from the North Pole
and go20mph due north by the compass.How long will it take to get to the Pole?Answer is that you never get there _ you get to the North
Magnetic Pole.

Ackermann.1925.P. 116.Man at North Pole
goes20miles south and30miles west.How far, and in what direction, is he from the Pole?

H.
Phillips.The Playtime Omnibus.Faber & Faber, London, 1933.Section XVI, prob. 11: Polar conundrum, pp.
51 & 234.Start at the North Pole,
go40miles South, then30miles West.How far are you from the Pole.Answer:"Forty miles.(NOT thirty, as one is tempted to
suggest.)"Thirty appears to be a
slip for fifty??

Jules
Leopold.At Ease!Op. cit. in 4.A.2.1943.A helluva
question!, pp. 10 & 196.Hunter
goes10mi south,10mi west, shoots a bear and drags it10mi back to his starting point.What colour was the bear?Says
the only geographic answer is the North Pole.

Northrop.Riddles in Mathematics.1944.1944: 5-6;1945: 5-6;1961: 15‑16.He starts with the house which faces south on all sides.Then he has a hunter that sees a bear 100
yards east.The hunter runs 100 yards
north and shoots south at the bear _ what colour ....He then gives the three‑sided walk version, but doesn't
specify the solution.

W. A.
Bagley.Puzzle Pie.Op. cit. in 5.D.5.1944.No. 50: A fine
outlook, pp. 54-55.House facing south
on all sides used by an artist painting bears!

Leeming.1946.Chap. 3, prob. 32: What color was the bear?, pp. 33 & 160.Man walks10miles south, then10miles west, where he shoots a bear.He drags it10miles north to his base.What color ....He gives only one solution.

Benjamin L.
Schwartz.What color was the
bear?.MM 34 (1960) 1-4.??NYS _ described by Gardner, SA (May 1966)
= Carnival, chap. 17.Considers the
problem where the hunter looks south and sees a bear100yards away.The bear goes100yards east and the
hunter shoots it by aiming due south.This gives two extra types of solution.

David
Singmaster.Bear hunting problems.Submitted to MM, 1986.Finds explicit solutions for the general
version of Perelman/Klamkin's problem.[In fact, I was ignorant of (or had long forgotten) the above when I
remembered and solved the problem.My
thanks to an editor (Paul Bateman ??check) for referring me to Klamkin.The Kakinuma et al then turned up
also.]Analysis of the solutions leads
to some variations, including the following.

David
Singmaster.Home is the hunter.Man heads north, goes ten miles, has lunch,
heads north, goes ten miles and finds himself where he started.

Bob Stanton.The explorers.Games Magazine 17:1 (No. 113) (Feb 1993) 61 & 43.Two explorers set out and go500miles in each direction.Madge
goes N, W, S, E, while Ellen goes E, S, W, N.At the end, they meet at the same point.However, this is not at their starting point.How come?and how far are they from their starting point, and in what
direction?They are not near either
pole.

Yuri B.
Chernyak & Robert S. Rose.The
Chicken from Minsk.BasicBooks, NY,
1995.Chap. 11, prob. 9: What color was
that bear? (A lesson in non-Euclidean geometry), pp. 97 &
185-191.Camper walks south 2 km, then
west 5 km, then north 2 km; how far is he from his starting point?Solution analyses this and related problems,
finding that the distancexsatisfies0 £
x £
7.183,noting that there are many
minimal cases near the south pole and if one is between them, one gets a local
maximum, so one has to determine one's position very carefully.

David
Singmaster.Symmetry saves the
solution.IN: Alfred S. Posamentier
& Wolfgang Schulz, eds.; The Art of Problem Solving: A Resource for the
Mathematics Teacher; Corwin Press, NY, 1996, pp. 273-286.Sketches the explicit solution to Klamkin's
problem as an example of the use of symmetric variables to obtain a solution.

Anonymous.Brainteaser B163 _ Shady matters.Quantum 6:3 (Jan/Feb 1996) 15 & 48.Is there anywhere on earth where one's
shadow has the same length all day long?

6.AG.MOVING
AROUND A CORNER

There are several versions of
this.The simplest is moving a ladder
or board around a corner _ here the problem is two-dimensional and the ladder
is thin enough to be considered as a line.There are slight variations _ the corner can be at aTor+junction;the widths of
the corridors may differ;the angle may
not be a right angle;etc.If the object being moved is thicker _ e.g.
a table _ then the problem gets harder.If one can use the third dimension, it gets even harder.

H. E.
Licks.Op. cit. in 5.A, 1917. Art. 110, p. 89.Stick going into a circular shaft in the ceiling.Gets[h2/3 + d2/3)]3/2for maximum length, wherehis
the height of the room anddis the diameter of the shaft."A simple way to solve a problem which
has proved a stumbling block to many."

Abraham.1933.Prob. 82 _ Another ladder, pp. 37 & 45 (23 & 117).Ladder to go from one street to another, of
different widths.

J. S.
Madachy.Turning corners.RMM 5 (Oct 1961) 37,6 (Dec 1961) 61&8 (Apr 1962)
56.In 5, he asks for the greatest
length of board which can be moved around a corner, assuming both corridors
have the same width, that the board is thick and that vertical movement is
allowed.In 6, he gives a numerical
answer for his original values and asserts the maximal length for planar
movement, with corridors of widthwand plank of thicknesst,is2 (wÖ2 ‑
t).In vol. 8, he says no two solutions
have been the same.

L. Moser,
proposer;M. Goldberg and J. Sebastian,
solvers.Problem 66‑11 _ Moving
furniture through a hallway.SIAM
Review 8 (1966) 381‑382&11 (1969) 75‑78&12 (1970) 582‑586."What is the largest area region which can be moved through a
"hallway" of width one (see Fig. 1)?"The figure shows that he wants to move around a rectangular
corner joining two hallways of width one.Sebastian (1970) studies the problem for moving an arc.

J. M.
Hammersley.On the enfeeblement of
mathematical skills ....Bull. Inst.
Math. Appl. 4 (1968) 66‑85.Appendix IV _ Problems, pp. 83‑85, prob. 8, p. 84.Two corridors of width1at
a corner.Show the largest object one
can move around it has area< 2 Ö2 and
that there is an object of area³π/2 + 2/π=2.2074.

Research
news:Conway's sofa problem.Mathematics Review 1:4 (Mar 1991) 5-8 &
32.Reports on Joseph Gerver's almost
complete resolution of the problem in 1990.Says Conway asked the problem in the 1960s and that Moser is the first
to publish it.Says a grop at a
convexity conference in Copenhagen improved Hammersley's results to2.2164.Gerver's analysis gives an object made up of 18 segments with area2.2195.The analysis depends on some unproven general assumptions which seem
reasonable and is certainly the unique optimum solution given those
assumptions.

A. A.
Huntington.More on ladders.M500 145 (Jul 1995) 2-5.Does usual problem, geting a quartic.The finds the shortest ladder.[This turns out to be the same as the
longest ladder one can get around a corner from corridors of widthswandh,so 6.AG is related to 6.L.]

6.AH.TETHERED
GOAT

A goat is grazing in a circular field
and is tethered to a post on the edge.He can reach half of the field.How long is the rope?There are
numerous variations obtained by modifying the shape of the field or having
buildings within it.In recent years,
there has been study of the form where the goat is tethered to a point on a
circular silo in a large field _ how much area can he graze?

Ladies Diary,
1748.P. 41.??NYS

Dudeney.Problem 67: Two rural puzzles _ No. 67: One
acre and a cow.Tit‑Bits 33 (5
Feb&5 Mar 1898) 355&432.Circular field opening onto a small
rectangular paddock with cow tethered to the gate post so that she can graze
over one acre.By skilful choice of
sizes, he avoids the usual transcendental equation.

Arc. [R. A.
Archibald].Involutes of a circle and a
pasturage problem.AMM 28 (1921) 328‑329.Cites Ladies Diary and it appears that it
deals with a horse outside a circle.

J.
Pedoe.Note 1477:An old problem.MG 24 (No. 261) (Oct 1940) 286-287.Finds the relevant area by integrating in polar coordinates
centred on the post.

A. J.
Booth.Note 1561:On Note 1477.MG 25 (No. 267) (Dec 1941) 309‑310.Goat tethered to a point on the perimeter of
a circle which can graze over½, _,
¼of the area.

No.
8: "Don't fence me in", pp. 87.Equilateral triangular field of area 120.Three goats tethered to the corners with ropes of length equal to
the altitude.Consider an area
wherengoats graze as contributing1/nto each goat.What area does each goat graze over?

No.
53: Around the silo, pp. 71 & 112-113.Goat tethered to the outside of a silo of diameter 20 by a rope of
length10π,i.e. he can just get to the other side of
the silo.How big an area can he
graze?The curve is a semicircle
together with two involutes of a circle, so the solution uses some calculus.

Marshall
Fraser.A tale of two goats.MM 55 (1982) 221‑227.Gives examples back to 1894.

Bull, 1998,
below, says this problem has been discussed by the Internet newsgroupsci.mathsome years previously.

Michael E.
Hoffman.The bull and the silo: An
application of curvature.AMM 105:1
(Jan 1998) ??NYS _ cited by Bull.Bull
is tethered by a rope of lengthLto a circular silo of radiusR.IfL £ πR,then the grazeable area isL3/3R + πL2/2.This paper considers the problem for general
shapes.

John
Bull.The bull and the silo.M500 163 (Aug 1998) 1-3.Improves Hoffman's solution for the circular
silo by avoiding polar coordinates and using a more appropriate variable,
namely the angle between the taut rope and the axis of symmetry.

6.AI.TRICK
JOINTS

S&B, pp. 146‑147, show
several types.

These are often made in two contrasting
woods and appear to be physically impossible.They will come apart if one moves them in the right direction.A few have extra complications.The simplest version is a square cylinder
with dovetail joints on each face _ called common square version below.There are also cases where one thinks it
should come apart, but the wood has been bent or forced and no longer comes
apart _ see also 6.W.5.

Tom Tit,
vol. 2.1892.Assemblage paradoxal, pp. 231-232.= K, no. 155: The paradoxical coupling, pp. 353‑354.Common square version with instructions for
making it by cutting the corners off a larger square.

T.
Moore.A puzzle joint and how to make
it.The Woodworker 1:8 (May 1902)
172.S&B, p. 147, say this is the
earliest reference to the common square version."... the foregoing joint will doubtless be well-known to our
professional readers.There are
probably many amateur woodworkers to whom it will be a novelty."

Dudeney.The world's best puzzles.Op. cit. in 2.1908.Shows the common
square version "given to me some ten years ago, but I cannot say who first
invented it."He previously
published it in a newspaper.??look in
Weekly Dispatch.

Dudeney.AM.1917.Prob. 424: The dovetailed
block, pp. 145 & 249.Shows the
common square version _ "... given to me some years ago, but I cannot say
who first invented it."He
previously published it in a newspaper.??as above

Allan
Boardman.Up and Down Double
Dovetail.Shown on p. 147 of
S&B.Square version with alternate
dovetails in opposite directions.This
is impossible!

I have a set
of examples which belonged to Tom O'Beirne.There is a common square version and a similar hexagonal version.There is an equilateral triangle version
which requires a twist.There is a
right triangle version which has to be moved along a space diagonal![One can adapt the twisting method ton-gons!]

Dick
Schnacke (Mountain Craft Shop, American Ridge Road, Route 1, New Martinsville,
West Virginia, 26155, USA) makes a variant of the common square version which
has two dovetails on each face.I
bought one in 1994.

6.AJ.GEOMETRIC
ILLUSIONS

Anonymous
15C French illustrator of Giovanni Boccaccio, De Claris Mulieribus, MS Royal 16
Gv in the British Library.F. 54v:
Collecting cocoons and weaving silk.??NYS _ reproduced in: The Medieval WomanAn Illuminated Book of Postcards, HarperCollins, 1991.This shows a loom(?) frame with uprights at
each corner and the crosspieces joining the tops of the end uprights as though
front and rear are reversed compared to the ground.

L. A.
Necker.LXI. Observations on some
remarkable optical phnomena seen in Switzerland; and on an optical phnomenon
which occurs on viewing a figure of a crystal or geometrical solid.Phil. Mag. (3) 1:5 (Nov 1832) 329-337.This is a letter from Necker, written on 24
May 1832.On pp. 336-337, Necker
describes the visual reversing figure known as the Necker cube which he
discovered in drawing rhomboid crystals.This is also quoted in Ernst; The Eye Beguiled, pp. 23-24].Richard L. Gregory [Mind in Science;
Weidenfeld and Nicolson, London, 1981, pp. 385 & 594] and Ernst say that
this was the first ambiguous figure to be described.

Wehman.New Book of 200 Puzzles.1908.The cube puzzle, p. 37.A 'baby
blocks' pattern of cubes, which appears to show six cubes piled in a corner one
way and seven cubes the other way.I
don't recall seeing this kind of puzzle in earlier sources??

Lietzmann,
Walther&Trier, Viggo.Wo steckt
der Fehler?3rd ed., Teubner, 1923.[The Vorwort says that Trier was coauthor of
the 1st ed, 1913, and contributed most of the Schülerfehler (students'
mistakes).He died in 1916 and
Lietzmann extended the work in a 2nd ed of 1917 and split it into Trugschlüsse
and this 3rd ed.There was a 4th ed.,
1937.See Lietzmann for a later version
combining both parts.]II. Täuschungen
der Anschauung, pp. 7-13.

Williams.Home Entertainments.1914.Colour discs for the gramophone, pp. 207-212.Discusses several effects produced by spirals and eccentric
circles on discs when rotated.

Gerald H.
Fisher.The Frameworks for Perceptual
Localization. Report of MOD Research Project70/GEN/9617, Department of
Psychology, University of Newcastle upon Tyne, 1968.Good collection of examples, with perhaps the best set of
impossible figures.

Pp. 42‑47 _ reversible
perspectives.

Pp. 56‑65 _ impossible
and ambiguous figures.

Appendix 6, p.190 _ 18
reversible figures.

Appendix 7, pp. 191‑192
_ 12 reversible silhouettes.

Appendix 8, p. 193 _ 12
impossible figures.

Appendix 14, pp. 202‑203
_ 72 geometrical illusions.

Harvey
Long."It's All In How You Look At
It".Harvey Long & Associates,
Seattle, 1972.48pp collection of
examples with a few references.

J. R. Block
& Harold E. Yuker.Can You Believe
Your Eyes?Brunner/Mazel, NY,
1992.Excellent survey of the field of
illusions, classified into 17 major types _ e.g. ambiguous figures, unstable
figures, ..., two eyes are better than one.They give as much information as they can about the origins.They give detailed sources for the following
_ originals NYS??.These are also
available as two decks of playing cards.

W.
E. Hill.My wife and my
mother-in-law.Puck,
(6 Nov 1915) 11.[However,
Julian Rothenstein & Mel Gooding; The Paradox Box; Redstone Press, London,
1993; include a reproduction of a German visiting card of 1888 with a version
of this illusion.The English caption
by James Dalgety is:My Wife and my
Mother-in-law.Cf Seckel, below.]Ernst, just above, cites Hill and says he
was a cartoonist, but gives no source.Long, above, asserts it was designed by E. G. Boring, an American
psychologist.

Al
Seckel.Illusions in Art.Two decks of playing cards in case with
notes.Deck 1 _ Classics.Works from Roman times to the middle of the
20th Century.Deck 2 _
Contemporary.Works from the second
half of the 20th Century.Y&B
Associates, Hempstead, NY, 1997.This
gives further details on some of the classic illusions _ some of this is
entered above and in 6.AU and some is given below.

Hole location
gage.Analog Science Fact  Science
Fiction 73:4 (Jun 1964) 27.Classic Two
pronged trident, with some measurements given.Editorial note says the item was 'sent anonymously for some reason' and
offers the contributor $10 or a two year subscription if he identifies
himself.(Thanks to Peter McMullen for
the Analog items, but he doesn't recall the contributor ever being named.)

Edward G.
Robles, Jr.Letter (Brass Tacks
column).Analog Science Fact  Science
Fiction 74:4 (Dec 1964) 4.Says the Jun
1964 object is a "three-hole two slot BLIVIT" and was developed at
JPL (Jet Propulsion Laboratory, Pasadena) and published in their Goddard
News.He provides a six-hole five-slot
BLIVIT, but as the Editor comments, it 'lacks the classic simple elegance of
the Original.'

Mad
Magazine.No. 93 (Mar 1965).Cover.Miniature reproduction in:Maria
Reidelbach; Completely Mad _ A History of the Comic Book and Magazine; Little,
Brown & Co., Boston, 1991, p. 82.Shows a standard version.Al
Seckel says they thought it was an original idea and they apologised in the
next issue _ to whom??

Heinz Von
Foerster.From stimulus to symbol: The
economy of biological computation.IN:Sign Image Symbol;ed. Gyorgy Kepes;Studio Vista, London, 1966, pp. 42-60.On p. 55, he shows the "Triple-pronged fork with only
two branches" and on p. 54, he notes that although each portion is
correct, it is impossible overall, but he gives no indication of its history or
that it is at all new.

Pearson.1907.Part II, no. 3: Whirling wheels, p. 3.Gives Thompson's form, but the wheels are overlapping, which makes it
look a bit like an ancestor of the tribar.

Oscar
Reutersvård.Omöjliga Figure
[Impossible Figures _ In Swedish].Edited by Paul Gabriel.Doxa,
Lund, (1982);2nd ed., 1984.This seems to be the first publication of
his work, but he has been exhibiting since about 1960 and some of the
exhibitions seem to have had catalogues.P. 9 shows and discusses his Opus 1 from 1934, which is an impossible
tribar made from cubes.(Reproduced in
Ernst, 1992, p. 69 as a drawing signed and dated 1934.Ernst quotes Reutersvård's correspondence
which describes his invention of the form while doodling in Latin class as a
schoolboy.A school friend who knew of
his work showed him the Penroses' article in 1958 _ at that time he had drawn
about 100 impossible objects _ by 1986, he had extended this to some
2500!)He has numerous variations on
the tribar and the two‑pronged trident.

Oscar
Reutersvård.Swedish postage stamps for
25, 50, 75 kr.1982, based on his
patterns from the 1930s.The 25 kr. has
the tribar pattern of cubes which he first drew in 1934.(Also the 60 kr.??)

L. S. &
R. Penrose.Impossible objects: A
special type of visual illusion.British Journal of Psychology 49 (1958) 31‑33.Presents tribar and staircase.Photo of model staircase.[Ernst, 1992, pp. 71-73, quotes conversation
with Penrose about his invention of the Tribar and reproduces this
article.Penrose, like the rest of us,
only learned about Reutersvård's work in the 1980s.]

Anon.(?)Don't believe it.Daily Telegraph (24 Mar 1958) ?? (clipping found in an old
book)."Three pages of the latest
issue of the British Journal of Psychology are devoted to "Impossible
Objects.""Shows both the
tribar and the staircase.

This is the illusion seen in
alternatingly coloured brickwork where the lines of bricks distinctly seem
tilted.I suspect it must be apparent
in brickwork going back to Roman times.

The illusion
is apparent in the polychrome brick work on the side wall inside Keble College
Chapel, Oxford, by William Butterfield, completed in 1876 [thanks to Deborah
Singmaster for observing this].

Lietzmann
& Trier, op. cit. at 6.AJ, 1923.Pp. 12-13 has a striking version of this, described as a 'Flechtbogen
der Kleinen'.I can't quite translate
this _ Flecht is something interwoven but Bogen could be a ribbon or an arch or
a bower, etc.They say it is reproduced
from an original by Elsner.See Lietzmann,
1953.

Richard A.
Proctor.Gossip column.Knowledge 10(Dec 1886) 43&(Feb 1887) 92.6 x 6array of
cells.Prisoner in one corner can exit
from the opposite corner if he passes "once, and once only, through all
the36cells.""... take the
prisoner into either of the cells adjoining his own, and back into his own,
....This puzzle is rather a sell,
...."Letter and response [in
Gossip column, Knowledge 10 (Mar 1887) 115-116] about the impossibility of any
normal solution.

Loyd.Problem 16: The captive maiden.Tit‑Bits 31 (30 Jan&20 Feb 1897) 325&381.Rook's tour in minimal number of moves from a corner to the diagonally
opposite corner, entering each cell once.Because of parity, this is technically impossible, so the first two
moves are into an adjacent cell and then back to the first cell, so that the
first cell has now been entered.

Loyd.Problem 20: Hearts and darts.Tit‑Bits 31 (20 Feb,13&20 Mar 1897) 381,437,455. Queen's tour on8 x 8,starting in a corner, permitting crossings, but with no segment going
through a square where the path turns.Solution in14segments.This is No. 41 in White _ see the first Loyd entry above.

Ball.MRE, 4th ed., 1905, p. 197.At the end of his section on knight's tours,
he states that there are many similar problems for other kinds of pieces.

Loyd.The postman's puzzle.Cyclopedia, 1914, pp. 298 & 379.Rook's circuit on8 x 8array of points,
with one point a bit out of line, starting and ending at a central square,
in16segments.P. 379 also shows
another8 x 8circuit, but with a slope2line.See also pp. 21 & 341 and SLAHP, pp. 85
& 118, for two more examples.

W. Leslie
Prout.Think Again.Frederick Warne & Co., London,
1958.Joining the stars, pp. 41 &
129.5 x 5array of points.Using a
line of four segments, pass through17points.This is a bit like the3 x
3problem in that one must go outside
the array.

André
Viricel (with Jacques Bouteloup).Le
Théorème de Morley.L'Association pour
le Développement de la Culture Scientifique, Amiens, 1993.[This publisher or this book was apparently
taken over by Blanchard as Blanchard was selling copies with his label pasted
over the previous publisher's name in Dec 1994.]A substantial book (180pp) on all aspects of the theorem.The bibliography is extremely cryptic, but
says it is abridged from Mathesis (1949) 175??NYS.The most recent item
cited is 1970.

6.AN.VOLUME
OF THE INTERSECTION OF TWO CYLINDERS

Archimedes.The Method:Preface, 2.In:T. L. Heath; The Works of Archimedes, with a
supplement "The Method of Archimedes"; (originally two works, CUP,
1897 & 1912)= Dover, 1953.Supplement, p. 12, states the result.The proof is lost, but pp. 48‑51
gives a reconstruction of the proof by Zeuthen.

Liu
Hui.Jiu Zhang Suan Chu Zhu (Commentary
on the Nine Chapters of the Mathematical Art).263.??NYS _ described in Li
& Du, pp. 73‑74 & 85.He
shows that the ratio of the volume of the sphere to the volume of Archimedes'
solid, called mou he fang gai (two square umbrellas), isπ/4,but he cannot determine either volume.

He considers the shape,
called fanggai, within the natural circumscribed cube and shows that, in each
octant, the part of the cube outside the fanggai has cross section of areah2at distancehfrom the centre.This is equivalent to a tetrahedron, whose volume had been
detemined by Liu, so the excluded volume is_of the cube.

Li & Du, pp. 85‑87,
and say the result may have been found c480 by Zu Geng's father, Zu Chongzhi.

Lam Lay-Yong
& Shen Kangsheng.The Chinese
concept of Cavalieri's Principle and its applications.HM 12 (1985) 219-228.Discusses the work of Liu and Zu.

Shiraishi
Ch_ch_.Shamei Sampu.1826.??NYS _ described in Smith & Mikami, pp. 233-236."Find the volume cut from a cylinder by
another cylinder that intersects is orthogonally and touches a point on the
surface".I'm not quite sure what
the last phrase indicates.The book
gives a number of similar problems of finding volumes of intersections.

P. R. Rider,
proposer;N. B. Moore, solver.Problem 3587.AMM 40 (1933) 52 (??NX) &612.Gives the standard proof by
cross sections, then considers the case of unequal cylinders where the solution
involves complete elliptic integrals of the first and second kinds.References to solution and similar problem
in textbooks.

Trick
versions _ with doubled counters:Family Friend (1858),Illustrated Boy's Own Treasury,Secret Out,Hoffmann
(1876),Mittenzwey,Hoffmann (1893), nos. 8 & 9,Pearson,Home Book ....These could also
be considered as in 7.Q.2 or 7.Q.

Family
Friend (Dec 1858) 359.Practical
puzzles _ 2."Make a square with
twelve counters, having five on each side."(12, 4, 5).I haven't got
the answer, but presumably it is the trick version of a hollow square with
doubled corners, as in 7.Q.See
Illustrated Boy's Own Treasury, 1860.

Book of 500
Puzzles.1859.Prob.3, 9, 12, 17, 22, 26, 27, 29, 32, 36are identical to those in The Sociable, with page numbers decreased by
282.

Prob.
13, pp. 397 & 438."Make a
square with twelve counters, having five on each side."(12, 4, 5).Trick version of a hollow square with doubled corners.Presumably identical to Family Friend,
1858.Same as Secret Out.

Magician's
Own Book (UK version).1871.The solution to The florist's puzzle (The
Sociable, prob. 8) is given at the bottom of p. 284, apparently to fill
out the page as there is no relevant text anywhere.

Hanky
Panky.1872.

To
place nine cards in ten rows of three each, p. 291.I.e.(9, 10, 3).

Diagram
with no text, p. 128.(37, 20, 5),equally spaced on each line as in The
Sociable, prob. 27.

Hoffmann.Modern Magic.(George Routledge, London, 1876);reprinted by Dover, 1978.To place twelve cards in rows, in such a manner that they will count
four in every direction, p. 58.Trick
version of a3 x 3square with extras on a diagonal, giving a
form of(12, 7, 4).

Given
two rows of five dots, move four to make 5 rows of 4.Shulman describes this case, following Dudeney, AM, 1917, then
observes that since Dudeney is using coins, there are further solutions by
putting a coin on top of another.He
refers to Hoffmann and Loyd.

(9,
10, 3).Shulman quotes from Robert T.
Philip; Family Pastime; London, 1852, p. 30, ??NYS, but this must refer to the
item in Family Friend, which was edited by Robert Kemp Philp.BMC indicates Family Pastime may be another
periodical.Shulman then cites Jackson
and Dudeney.

Mittenzwey.1879?

Prob.
174, pp. 33 & 82.(6, 3, 3).(6, 4, 3)by a trick.

Prob.
175, pp. 33 & 82.Arrange 16
pennies as a3 x 3square so each row and column has four in
it.Solution shows a3 x 3square with extras on the diagonal _ but this only uses 12 pennies!So this the trick version of(12, 7, 4)as in Hoffmann (1876).

Prob.
176, pp. 33 & 82.(21, 10, 5).

Cassell's.1881.P. 92: The six rows puzzle.=
Manson, 1911, p. 146.

J. J.
Sylvester.Problem 2572.Math. Quest. from the Educ. Times 45 (1886)
127‑128.??NYS _ cited in Burr,
below.Obtains goodexamples of(a, b, 3)for eacha.In most cases, this is still the best known.

Dudeney."The Captain" puzzle corner.The Captain 3:2 (May 1900) 179.This gives a solution of a problem called Joubert's
guns, but I haven't seen the proposal.(10, 5, 4)but wants the maximum
number of castles to be inside the walls joining the castles.Manages to get two inside.= Dudeney; The puzzle realm; Cassell's
Magazine ?? (May 1908) 713-716; no. 6: The king and the castles.= AM, 1917, prob. 206: The king and the
castles, pp. 56 & 189.

Part
II, no. 83: For the children, pp. 83 & 177.Trick version of(12, 4,
5),as in Family Friend (1858).

Dudeney.The world's best puzzles.Op. cit. in 2.1908.He says(9, 10, 3)"is attributed to Sir Isaac Newton, but the earliest collection of
such puzzles is, I believe, in a rare little book that I possess _ published in
1821."[This must refer to Jackson.]Says Rev. Mr. Wilkinson gave(11, 16, 3)"some quarter of a century ago" and that he, Dudeney,
published(16, 15, 4)in 1897 (cf under 1902 above).He leaves these as problems but doesn't give
their solutions in the next issue.

Prob.
210: The ten coins.Two rows of
five.Move four to make(10, 5, 4).Cf. Carroll, 1876.Shows there
are 2400 ways to do this.He shows that
there are six basic solutions of the(10, 5, 4)which he calls:star, dart, compasses, funnel, scissors,
nail and he describes the smallest arrays on which they can fit.

Prob.
211: The twelve mince-pies.12 points
at the vertices and intersections of a Star of David.Move four to make(12, 7,
4).

R.
Ripley.Believe It or Not!Book 2.Op. cit. in 5.E, 1931.The
planter's puzzle, p. 197, asks for(19,
9, 5)but no solution is given.See Clark, above, for a better version of
the verse.

"I am constrained to plant a
grove

For a lady that I love.

This ample grove is too composed;

Nineteen trees in nine
straight rows.

Five trees in each row I must
place,

Or I shall never see her face."

Rudin.1936.Nos. 105-108, pp. 39 & 99-100.

No.
105: (9, 10, 3).

No.
106: (10, 5, 4)_ two solutions.

No.
107: (12, 6, 4)_ two solutions.

No.
108: (19, 9, 5).

Depew.Cokesbury Game Book.1939.The orange grower, p. 221.(21,
9, 5).

The Home
Book of Quizzes, Games and Jokes.Op.
cit. in 4.B.1, 1941.P. 147, prob. 1
& 2.Place six coins in anLor
a cross and make two rows of four, i.e.(6, 2, 4),which is
done by the simple trick of putting a coin on the intersection.

Loyd.Problem 34: War‑ships at anchor.Tit‑Bits 32 (22 May&12 Jun 1897) 135&193.Place four warships equidistantly so that if one is attacked, the others
can come to assist it.Solution is a
tetrahedron of points on the earth's oceans.

Prob.
564-31, pp. 254 & 396.From a6 x 6array, remove6to leave an even number in each row.(The German 'Reihe' can be interpreted as
row or column or both.)If we consider
this in the first quadrant with coordinates going from 1 to 6, the removed
points are:(1,2), (1,3), (2,1), (2,2),
(6,1), (6,3).The use of the sixth
column is peculiar and has the effect of making both diagonals odd, while the
more usual use of the third column would make both diagonals even.

Prob.
583-5, pp. 285 & 403: Von folgenden 36 Punkten sechs zu streichen.As above, but each file ('Zeile') in 'all
four directions' has four or six points.Deletes:(1,1), (1,2),
(2,2), (2,3), (6,1), (6,3)which makes
one diagonal even and one odd.

Mittenzwey.1879?Prob. 177, pp. 33 & 82.Given
a4 x 4array, remove 6 to leave an even number in each row and
column.Solution removes a2 x 3rectangle from a corner.

Hoffmann.1893.

Chap.
VI, no. 22: The thirty‑six puzzle, pp. 271 & 285.Place 30 counters on a6 x 6board so each horizontal and each vertical line has an even number.Solution places the six blanks in a3 x 3corner in the obvious way.This
also makes the diagonals have even numbers.

Chap.
VI, no. 23: The "Five to Four" puzzle, pp. 272 & 285.Place 20 counters on a5 x 5board subject to the above conditions.Solution puts blanks on the diagonal.This also makes the diagonals have even number.

Dudeney.The puzzle realm.Cassell's Magazine ?? (May 1908) 713-716.The crack shots.10 pieces in a4 x 4array making the maximal number of even
lines _ counting diagonals and short diagonals _ with an additional
complication that pieces are hangine in in vertical strings.The picture is used in AM, prob. 270.

Obermair.Op. cit. in 5.Z.1.1984.Prob. 37, pp. 38
& 68.52 men on an8 x 8board with all rows, columns and diagonals (both long and short) having
an even number.

6.AP.DISSECTIONS
OF A TETRAHEDRON

6.AP.1.TWO PIECES

Richard A.
Proctor.Our puzzles;Knowledge 10 (Feb 1887) 83&Solutions of puzzles;Knowledge
10 (Mar 1887) 108-109."Puzzle
XIX.Show how to cut a regular
tetrahedron (equilateral triangular pyramid) so that the face cut shall be a
square: also show how to plug a square hole with a tetrahedron."Solution shows the cut clearly.

Donovan A.
Johnson.Paper Folding for the
Mathematics Class.NCTM, 1957, p. 26:
Pyramid puzzle.Gives instructions for
making the pieces from paper.

Claude
Birtwistle.Editor's footnote.MTg 21 (Winter 1962) 32."The following interesting puzzle was
given to us recently."

Birtwistle.Math. Puzzles & Perplexities.1971.Bisected tetrahedron, pp. 157-158.Gives the net so one can make a drawing, cut it out and fold it up to
make one piece.

6.AP.2.FOUR PIECES

These dissections usually also work
with a tetrahedron of spheres and hence these are related to ball pyramid
puzzles, 6.AZ.

The first version I had in mind
dissects each of the two pieces of 6.AP.1 giving four congruent rhombic
pyramids.Alternatively, imagine a
tetrahedron bisected by two of its midplanes, where a midplane goes halfway
between a pair of opposite edges.This
puzzle has been available in various versions since at least the 1970s,
including one from Stokes Publishing Co., 1292 Reamwood Avenue, Sunnyvale,
California, 94089, USA., but I have no idea of the original source.The same pieces are part of a more complex
dissection of a cube, PolyPackPuzzle, which was produced by Stokes in
1996.(I bought mine from Key
Curriculum Press.)

In 1997, Bill Ritchie, of Binary Arts,
sent a quadrisection of the tetrahedron that they are producing.Each piece is a hexahedron.The easiest way to describe it is to consider
the tetrahedron as a pile of spheres with four on an edge and hence20altogether.Consider a planar
triangle of six of these spheres with three on an edge and remove one vertex
sphere to produce a trapezium (or trapezoid) shape.Four of these assemble to make the tetrahedron.Writing this has made me realise that Ray
Bathke has made and sold these5-spherepieces as Pyramid 4 for
a few years.However, the solid pieces
used by Binary Arts are distinctly more deceptive.

Len Gordon produced another quadrisection of the20sphere tetrahedron 00

using the planar shape at the
right.This was c1980??000

David
Singmaster.Sums of squares and
pyramidal numbers.MG 66 (No. 436) (Jun
1982) 100-104.Consider a tetrahedron
of spheres with2non an edge.The quadrisection described above gives four pyramids whose layers are
the squares1, 4, ..., n2.Hencefour times the sum of the firstnsquares is the tetrahedral
number for2n,i.e. 4 [1 + 4 + ...
+ n2] = BC(n+2, 3).

6.AQ.DISSECTIONS
OF A CROSS,TORH

The usual dissection of a cross has
two diagonal cuts at45oto the sides and passing through two of the
reflex corners of the cross and yielding five pieces.The central piece is six-sided, looking like a rectangle with its
ends pushed in.Depending on the
relative lengths of the arms, head and upright of the cross, the other pieces
may be isosceles right triangles or right trapeziums.Removing the head of the cross gives the usual dissection of theTinto four pieces _ then the central piece is five-sided.Sometimes the central piece is split in
halves.Occasionaly the angle of the
cuts is different than45o.Dissections of anHhave the same basic
idea of using cuts at45o_ the result can be a bit like twoTswith overlapping stems and the number of pieces depends on the relative
size and positioning of the crossbar of theH_ see: Rohrbough.

S&B, pp.
20‑21, show several versions.They say that crosses date from early 19C.They show a 6‑piece Druid's Cross, by Edwards & Sons,
London, c1855.They show severalT‑puzzles _ they say the first is an
1903 advertisement for White Rose Ceylon Tea, NY _ but see 1898 below.They also show someH‑puzzles.

Charles
Babbage.The Philosophy of Analysis _
unpublished collection of MSS in the BM as Add. MS 37202, c1820.??NX.See 4.B.1 for more details.F. 4
is "Analysis of the Essay of Games".F. 4.v has a cross cut into 5 pieces in the usual way.

Leske.Illustriertes Spielbuch für Mädchen.1864?Prob. 584-12, pp. 288 & 406: Ein Kreuz.Begins as the usual five piece cross, but the central piece is
then bisected into two mitres and the base has two bits cut off to give an
eight piece puzzle.

Lash, Inc. _
Clifton, N.J. _ Chicago, Ill. _ Anaheim, Calif.TPuzzle.Copyright Sept. 1898.4‑pieceTpuzzle to be cut out
from a paper card, but the angle of the cuts is about35oinstead
of45owhich makes it less symmetric and less
confusing than the more common version.The resultingTis somewhat wider than usual, being
about16%wider than it is tall.It
advertises:Lash's BittersThe Original Tonic Laxative.Photocopy sent by Slocum.

Rohrbough.Puzzle Craft.1932.The "H"
Puzzle, p. 23.Very squareH_
consider a3 x 3board with the top and bottom middle cells
removed.Make a cut along the main
diagonal and two shorter cuts parallel to this to produce four congruent
isosceles right triangles and two odd pentagons.

See
Rohrbough in 6.AS.1 for a very differentTpuzzle.

6.AR.QUADRISECTED
SQUARE PUZZLE

This is usually done by two
perpendicular cuts through the centre.A dissection proof of the Theorem of Pythagoras described by Henry
Perigal (Messenger of Mathematics 2 (1873) 104) uses the same shapes.For sidesa < b,the perpendicular cuts
are done in the square of sidebso they meet the sides at distance(b-a)/2from a corner.These pieces then
fit around the square of sideato make a square of sidec.Perigal is ??NYS, but described in Elisha Scott Loomis; The Pythagorean
Proposition; 2nd ed., NCTM, 1940, pp. 104-105 & 214, where some earlier
possible occurrences are mentioned.

The pieces make a number of other
different shapes.

Crambrook.1843.P. 4, no. 17: Four pieces to form a Square.This might be the dissection being considered here??

A.
Héraud.Jeux et Récréations
Scientifiques _ Chimie, Histoire Naturelle, Mathématiques.(1884);Baillière, Paris, 1903.Pp. 303‑304:
Casse‑tête.Uses two cuts which
are perpendicular but are not through the centre.He claims there are120ways to try to assemble it,
but his mathematics is shaky _ he adds the numbers of ways at each stage rather
than multiplying!Also, as Strens notes
in the margin of his copy (now at Calgary), if the crossing is off-centre, then
many of the edges have different lengths and the number of ways to try is
really only one.Actually, I'm not at
all sure what the number of ways to try is _ Héraud seems to assume one tries
each orientation of each piece, but some intelligence sees that a piece can
only fit one way beside another.

Handy Book
for Boys and Girls.Op. cit. in
6.F.3.1892.P. 14: The divided square puzzle.Crossing is off-centre.

Adams.Puzzle Book.1939.Prob. C.12: The
broken square, pp. 125 & 173.As
above, but notes that the pieces also make a square with a square hole.

6.AS.DISSECTION
OF SQUARES INTO A SQUARE

Lorraine
Mottershead.Investigations in
Mathematics.Blackwell, Oxford,
1985.P. 102 asserts that dissections
of squares to various hexagons and heptagons were known c1800 while square to
rectangle dissections were known to Montucla _ though she illustrates the
latter with examples like 6.Y, she must mean 6.AS.5.

6.AS.1.TWENTY 1, 2, Ö5 TRIANGLES MAKE A SQUARE

OR
FIVE EQUAL SQUARES TO A SQUARE

The basic puzzle has been varied in
many ways by joining up the 20 triangles into various shapes, but I haven't
attempted to consider all the modern variants.A common form is a square with a skew#in it, with each line joining
a corner to the midpoint of an opposite side, giving the 9 piece version.This has four of the squares having a
triangle cut off.For symmetry, it is
common to cut off a triangle from the fifth square, giving 10 pieces, though
the assembly into one square doesn't need this.See Les Amusemens for details.

If the dividing lines are moved a bit
toward the middle and the central square is bisected, we get a 10 piece puzzle,
having two groups of four equal pieces and a group of two equal pieces, called
the Japan square puzzle.I have
recently noted the connection of this puzzle with this section, so there may be
other examples which I have not previously paid attention to _ see:Magician's Own Book,Book of 500 Puzzles,Boy's Own Conjuring Book,Illustrated Boy's Own Treasury,Landells,Hanky Panky,Wehman.

Les
Amusemens.1749.P. xxxii.Consider five2 x 2squares.Make a cut from a corner to the midpoint of an opposite side on each
square.This yields five1, 2, Ö5triangles and five pieces comprising three such triangles.The problem says to make a square from five
equal squares.So this is the 10 piece
version.

Vyse.Tutor's Guide.1771?Prob. 6, p. 317
& Key p. 357.2 x 10board to be cut into five pieces to make
into a square.Cut into a2 x 2square and four2, 4, 2Ö5triangles.

Ozanam‑Montucla.1778.Avec cinq quarrés égaux, en former un seul.Prob. 18 & fig. 123, plate 15, 1778: 297;1803: 292-293;1814: 249-250;1840:
127.9 piece version.Remarks that any number of squares can be
made into a square _ see 6.AS.5.

Catel.Kunst-Cabinet.1790.

Das
mathematische Viereck, pp. 10-11 & fig. 15 on plate I.10 piece version with solution shown.Notes these make five squares.

Bestelmeier.1801.Item 629: Die 5 geometrisch zerschnittenen Quadrate, um aus 5 ein
einziges Quadrat zu machen.As in Les
Amusemens.S&B say this is the
first appearance of the puzzle.Only
shown in a box with one small square visible.

Minguét.Engaños.1822.Pp. 145-146.Not in 1733 or 1755 eds.9piece version.Also a 15piece version where triangles are cut off diagonally opposite corners of
each small square leaving parallelogram pieces as in Guyot.

Manuel des
Sorciers.1825.Pp. 201-202, art. 18.??NXFive squares to one _ usual 10 piece form and 15 piece form as in Guyot.

Nuts to
Crack IV (1835), no. 195.20 triangles
_ part of a long section: Tricks upon Travellers.The problem is used as a wager and the smart-alec gets it wrong.

The
Riddler.1835.The square of triangles, p. 8.Identical to Boy's Own Book, but without
illustration, some consequent changing of the text, and omitting the last
comment.

Crambrook.1843.P. 4.

No.
7: Egyptian Puzzle.Probably the 10
piece version as in Les Amusemens.See
S&B below, late 19C.Check??

No.
23: Twenty Triangles to form a Square.Check??

Boy's
Treasury.1844.Puzzles and paradoxes, no. 5, pp. 425 &
429."Cut twenty triangles out of
ten square pieces of wood;" and make a square.The solution shows that he means 'out of five square
pieces'.The phrasing is very similar
to Boy's Own Book.

Magician's
Own Book.1857.

How
to make five squares into a large one without any waste of stuff, p. 258.9 piece version.

Prob.
35: The Japan square puzzle, pp. 277 & 300.Make two parallel cuts and then two perpendicular to the first
two so that a square is formed in the centre.This gives a 9 piece puzzle, but here the central square is cut by a
vertical through its centre to give a 10 piece puzzle.= Landells, Boy's Own Toy-Maker, 1858, pp.
145-146.

Charles
Bailey (manufacturer in Manchester, Massachusetts).1858.An Ingenious Puzzle
for the Amusement of Children ....The
10 pieces of Les Amusemens, with 19 shapes to make, a la tangrams.Sent by Jerry Slocum _ it is not clear if
there were actual pieces with the printed material.

The
Sociable.1858.

Prob.
10: The protean puzzle, pp. 289 & 305-306.Cut a5 x 1into 11 pieces to form eight shapes, e.g. a
Greek cross.It is easier to describe
the pieces if we start with a10 x
2.Then three squares are cut off.One is halved into two1 x 2rectangles.Two squares have
two1, 2, Ö5triangles cut off leaving triangles of
sides2, Ö5, Ö5.The remaining double square is almost divided into halves each
with a1, 2, Ö5triangle cut off, but these two triangles
remain connected along their sides of size1,thus giving a4, Ö5, Ö5triangle and two trapeziums of sides2, 2, 1, Ö5.= Book of 500 Puzzles, 1859, prob. 10, pp. 7
& 23-24.